Ch3_LuoR

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= __**Projectile Motion**__ =

__**Shoot Your Grade (Rob K, Matt O, Ryan L) 11/4/11**__
__Video__ media type="file" key="Shoot Your Grade.mov" width="300" height="300" __Purpose and Hypothesis:__ Our group was trying to find where we should hang up the 5 rings so that the launched ball would go through all of them and where we should place the cup so the ball would land in it. We figured to find the location of the ball at a certain distance and hang a ring there, repeat this 4 more times, and find the location of where the ball would land and place the cup in that spot. We also thought that it would take several tries and a bit of good luck for the ball to pass through all the rings and land in our cup. This was based on our experiments from our previous lab, in which we determined that each launch was slightly different from the previous one due to a small shift of the launcher after each time we launched the ball, a different amount of power used each time we fire the launcher, and that the place where we set our cup was an averaged distance.

__Materials and Procedure:__ Everything provided for us in the box next to the launcher was used. Inside included the masking tape rings, duct tape, string, scissors, ball, carbon paper, and black tube. We used the string to tie the rings up against the ceiling and hung them down so that the height we calculated was the center of the ring. Afterwards, we calculated the total distance from where the ball was and launched to a little before where it landed as the place to put the cup. We placed the cup a little in front of our averaged distance in order to account for the height of the ball and it clearing the side of the cup. The total distance traveled by the ball was first marked by putting carbon paper on the ground and measuring the distance from the launcher to where the ball hit the paper. As there where several launches done to see the general area of where the ball would land, the final distance was an average of the distances to where the ball landed. We used this distance, the calculated height of where our ball was launched, angle at which our ball was launched (15º), and acceleration of the ball (0 for horizontal component, -9.8 m/s 2 for vertical component) to calculate the time it took for the ball to arrive at the destination and the initial velocity. We then used this information to find the position of the ball at a certain distance away from the launcher. For our calculations of where to hang up the rings, we used the formula ∆d=v o t+(1/2)at 2.

__Measurements:__ Distance was found by placing carbon paper in the general area of where the launched ball would land and measured the distance from where the ball was launched to where the ball landed. The 11.5 cm is added because it is the horizontal distance between where the ball is launched and the edge of the table, from which we measured out the initial distance starting at the 10 cm mark on the measuring tape, hence the -10 cm in the table. All in all, we added 1.5 cm to the initial distance to find our total distance.
 * Horizontal Displacement (x)**
 * Launch # || Distance from Edge of Table to Spot of Landing || Measurement adjustments || Final Distance ||
 * 1 || 4.870m || +11.5 cm - 10 cm || 4.885 m ||
 * 2 || 4.838m || +11.5 cm - 10 cm || 4.853 m ||
 * 3 || 4.793m || +11.5 cm - 10 cm || 4.808 m ||
 * 4 || 4.805m || +11.5 cm - 10 cm || 4.820 m ||
 * 5 || 4.811m || +11.5 cm - 10 cm || 4.826 m ||
 * 6 || 4.837m || +11.5 cm - 10 cm || 4.852 m ||
 * 7 || 4.810m || +11.5 cm - 10 cm || 4.825 m ||
 * Average ||  ||   || 4.838 m ||

Height from floor to where ball was launched- 1.172 m
 * Vertical Displacement (y)**

Green represents what was unknown
 * Initial Velocity**
 * || x || y ||
 * v o : || vcos(15º) || vsin(15º) ||
 * a: || 0 m/s 2 || -9.8 m/s 2 ||
 * t: || t= .71 s || t= .71 s ||
 * d: || 4.838 m || -1.172 m ||

X ∆d=v o t+(1/2)at 2 4.838=vcos(15º)t v=4.838/(cos(15º)t)

Y ∆d=v o t+(1/2)at 2 -1.172=vsin(15º)t - 4.9t 2

Substitute v equation calculated from the X equation into the V in the Y equation -1.172=(4.838/(cos(15º)t)(sin(15º)t - 4.9t 2 t=0.71 s

Substitute t into X equation V=4.838/(cos(15º)t) V=4.838/(cos(15º)(0.71)) V=7.054 m/s

cup height= .115 m Total change in distance: -1.172 m + .115 cm = -1.057 m (done to account for the ball clearing the side of the cup) Again, green represent originally unknown values.
 * Calculation on Where to Place the Cup**
 * || x || y ||
 * v o: || xcos(15º)=6.814 || xsin(15º)=1.826 ||
 * a: || 0 m/s 2 || -9.8 m/s 2 ||
 * t: || t= .687 s || t= .687 s ||
 * d: || d= 4.68 m || -1.057 m ||

To find the horizontal distance, we need to find the time it takes for the ball to reach the destination first. ∆d=v o t+(1/2)at 2 -1.057=1.826t - 4.9t 2 t=0.687 seconds

Substitute t into the X equation ∆d=v o t+(1/2)at 2 d=(0.687) x (6.814) d=4.68 m So we place the cup 4.68 meters away from where the ball is launched.

Ring 1: 1.10 m away from launching point Green represents unknown values
 * Calculations for Height of the Rings (1 example shown)**
 * || x || y ||
 * v o || 6.814 || 1.826 ||
 * a || 0 m/s 2 || -9.8 m/s 2 ||
 * t || t= .1614 s || t= .1614 s ||
 * d || 1.10 m || d= .1671 m ||

In order to find the vertical displacement, the time needed to arrive at destination must be calculated first ∆d=v o t+(1/2)at 2 1.10=6.814t t=0.1614 s

Substitute t into the Y equation to calculate vertical distance: ∆d=v o t+(1/2)at 2 d=(1.826)(.1614) - 4.9(.1614) 2 d=.1671 m (this distance represents how many meters the ring should be put above the launch height if positive or below the launch height if negative) So add .1671 m + 1.172 = 1.3391 m off the ground, or 133.91 cm off the ground.


 * Complete Calculations for Cup and Rings:**
 * Ring # || Time (s): || x (horizontal distance from launching point): || y (vertical distance from ground level): ||
 * 1 || .1614 || 1.10 m || 133.91 cm ||
 * 2 || .2539 || 1.73 m || 131.97 cm ||
 * 3 || .3317 || 2.26 m || 123.88 cm ||
 * 4 || .3977 || 2.71 m || 112.32 cm ||
 * 5 || .4726 || 3.22 m || 94.07 cm ||
 * Cup || .6870 || 4.68 m || 11.50 cm ||

Ring 1 (y value) [(theoretical-actual)/theoretical](100) [(1.3391-1.3391)/1.3391](100)=0% 0% error.
 * % Error Calculations:**

Ring 1 (x value) [(theoretical-actual)/theoretical](100) [(1.1-1.1)/1.1](100)=0 0% error

Ring 2 (y) (T-A)/T x 100 [(131.97-131.97)/131.97](100)=0 0% error

Ring 2 (x)


 * Complete % Error Calculations**
 * < Ring # ||< % Error (x) ||< % Error (y) ||
 * < 1 ||< 0% ||< 0% ||
 * < 2 ||< 0% ||< 0% ||
 * < 3 ||< 0% ||< 0% ||
 * < 4 ||< 0% ||< 0% ||
 * < 5 ||< 0% ||< 0% ||

__Conclusion:__ All in all, our hypothesis proved to be correct based on our calculations and the result of our experiment. At different distances, we calculated the height of the ball at those places, and placed a ring at the supposed location. At the end location, we placed the cup. Our hypothesis was proven correct when we succeeded in launching the ball into the cup the first time the ball went through all the rings. Our hypothesis was also supported by the fact that our experiment did not pass through all the rings during our first test, as a shift in power sent the ball powering against the edge of a ring of tape instead of through the air and into our cup.

While our error % is 0, we needed to adjust make sure none of our objects besides our ball moved during experimentation. We found out that the entire launcher shifted slightly after every launch, so we used duct tape to hold down the launcher. We also made sure that the angle at which we were firing our ball at did not change, as the degree sometimes increased after a launch. For our rings, our measurements for the height would be at the center of the ring so a small shift in the launch would not hit the side of a ring and continue its trajectory.

__**10/12/11**__
__Lesson 1 A & B__ A) **Vectors and Direction** A study of motion will involve the introduction of a variety of quantities that are used to describe the physical world. Examples of such quantities include distance, displacement, speed, velocity, acceleration, force, mass, momentum, energy, work, power, etc. All these quantities can by divided into two categories - [|vectors and scalars]. A vector quantity is a quantity that is fully described by both magnitude and direction. On the other hand, a scalar quantity is a quantity that is fully described by its magnitude. The emphasis of this unit is to understand some fundamentals about vectors and to apply the fundamentals in order to understand motion and forces that occur in two dimensions. Examples of vector quantities that have been [|previously discussed] include [|displacement], [|velocity], [|acceleration], and [|force]. Each of these quantities are unique in that a full description of the quantity demands that both a magnitude and a direction are listed. For example, suppose your teacher tells you "A bag of gold is located outside the classroom. To find it, displace yourself 20 meters." This statement may provide yourself enough information to pique your interest; yet, there is not enough information included in the statement to find the bag of gold. The displacement required to find the bag of gold has not been fully described. On the other hand, suppose your teacher tells you "A bag of gold is located outside the classroom. To find it, displace yourself from the center of the classroom door 20 meters in a direction 30 degrees to the west of north." This statement now provides a complete description of the displacement vector - it lists both magnitude (20 meters) and direction (30 degrees to the west of north) relative to a reference or starting position (the center of the classroom door). Vector quantities are not fully described unless both magnitude and direction are listed. Vector quantities are often represented by scaled [|vector diagrams]. Vector diagrams depict a vector by use of an arrow drawn to scale in a specific direction. Vector diagrams were introduced and used in earlier units to depict the forces acting upon an object. Such diagrams are commonly called as [|free-body diagrams]. An example of a scaled vector diagram is shown in the diagram at the right. The vector diagram depicts a displacement vector. Observe that there are several characteristics of this diagram that make it an appropriately drawn vector diagram.
 * a scale is clearly listed
 * a vector arrow (with arrowhead) is drawn in a specified direction. The vector arrow has a //head// and a //tail//.
 * the magnitude and direction of the vector is clearly labeled. In this case, the diagram shows the magnitude is 20 m and the direction is (30 degrees West of North).

**Conventions for Describing Directions of Vectors** Vectors can be directed due East, West, South, and North. But some vectors are directed //northeast// (45º angle); while others are more towards north than east (between 45º and 90º). Thus, there is a clear need for some form of a convention for identifying the direction of a vector that is __not__ due East, due West, due South, or due North. There are a variety of conventions for describing the direction of any vector. The two conventions that will be discussed and used in this unit are described below: > Two illustrations of the second convention (discussed above) for identifying the direction of a vector are shown below.
 * 1) The direction of a vector is often expressed as an angle of rotation of the vector about its "[|tail]" from east, west, north, or south. For example, a vector can be said to have a direction of 40 degrees North of West (meaning a vector pointing West has been rotated 40 degrees towards the northerly direction) of 65 degrees East of South (meaning a vector pointing South has been rotated 65 degrees towards the easterly direction).
 * 2) The direction of a vector is often expressed as a counterclockwise angle of rotation of the vector about its "[|tail]" from due East. Using this convention, a vector with a direction of 30 degrees is a vector that has been rotated 30 degrees in a counterclockwise direction relative to due east. A vector with a direction of 160 degrees is a vector that has been rotated 160 degrees in a counterclockwise direction relative to due east. A vector with a direction of 270 degrees is a vector that has been rotated 270 degrees in a counterclockwise direction relative to due east. This is one of the most common conventions for the direction of a vector and will be utilized throughout this unit.

Observe in the first example that the vector is said to have a direction of 40 degrees. You can think of this direction as follows: suppose a vector pointing East had its [|tail] pinned down and then the vector was rotated an angle of 40 degrees in the counterclockwise direction. Observe in the second example that the vector is said to have a direction of 240 degrees. This means that the tail of the vector was pinned down and the vector was rotated an angle of 240 degrees in the counterclockwise direction beginning from due east. A rotation of 240 degrees is equivalent to rotating the vector through two quadrants (180 degrees) and then an additional 60 degrees //into the// third quadrant.



**Representing the Magnitude of a Vector** The magnitude of a vector in a scaled vector diagram is depicted by the length of the arrow. The arrow is drawn a precise length in accordance with a chosen scale. For example, the diagram at the right shows a vector with a magnitude of 20 miles. Since the scale used for constructing the diagram is __1 cm = 5 miles__, the vector arrow is drawn with a length of 4 cm. That is, 4 cm x (5 miles/1 cm) = 20 miles. Using the same scale (__1 cm = 5 miles__), a displacement vector that is 15 miles will be represented by a vector arrow that is 3 cm in length. Similarly, a 25-mile displacement vector is represented by a 5-cm long vector arrow. And finally, an 18-mile displacement vector is represented by a 3.6-cm long arrow. See the examples shown below.



In conclusion, vectors can be represented by use of a scaled vector diagram. On such a diagram, a vector arrow is drawn to represent the vector. The arrow has an obvious tail and arrowhead. The magnitude of a vector is represented by the length of the arrow. A scale is indicated (such as, 1 cm = 5 miles) and the arrow is drawn the proper length according to the chosen scale. The arrow points in the precise direction. Directions are described by the use of some convention. The most common convention is that the direction of a vector is the counterclockwise angle of rotation which that vector makes with respect to due East.

B) **Vector Addition** A variety of mathematical operations can be performed with and upon vectors. One such operation is the addition of vectors. Two vectors can be added together to determine the result (or resultant). This process of adding two or more vectors has already been discussed in [|an earlier unit]. Recall in our discussion of Newton's laws of motion, that the //net force// experienced by an object was determined by computing the vector sum of all the individual forces acting upon that object. That is the [|net force] was the result (or [|resultant]) of adding up all the force vectors. During that unit, the rules for summing vectors (such as force vectors) were kept relatively simple. Observe the following summations of two force vectors:

These rules for summing vectors were applied to [|free-body diagrams] in order to determine the net force (i.e., the vector sum of all the individual forces). Sample applications are shown in the diagram below. In this unit, the task of summing vectors will be extended to more complicated cases in which the vectors are directed in directions other than purely vertical and horizontal directions. For example, a vector directed up and to the right will be added to a vector directed up and to the left. The //vector sum// will be determined for the more complicated cases shown in the diagrams below. There are a variety of methods for determining the magnitude and direction of the result of adding two or more vectors. The two methods that will be discussed in this lesson and used throughout the entire unit are:
 * the Pythagorean theorem and trigonometric methods
 * [|the head-to-tail method using a scaled vector diagram]

**The Pythagorean Theorem** The Pythagorean theorem is a useful method for determining the result of adding two (and only two) vectors __that make a right angle__ to each other. The method is not applicable for adding more than two vectors or for adding vectors that are __not__ at 90-degrees to each other. The Pythagorean theorem is a mathematical equation that relates the length of the sides of a right triangle to the length of the hypotenuse of a right triangle. To see how the method works, consider the following problem: >> Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Eric's resulting displacement. This problem asks to determine the result of adding two displacement vectors that are at right angles to each other. The result (or resultant) of walking 11 km north and 11 km east is a vector directed northeast as shown in the diagram to the right. Since the northward displacement and the eastward displacement are at right angles to each other, the Pythagorean theorem can be used to determine the resultant (i.e., the hypotenuse of the right triangle). The result of adding 11 km, north plus 11 km, east is a vector with a magnitude of 15.6 km. [|Later], the method of determining the direction of the vector will be discussed. Let's test your understanding with the following two practice problems. In each case, use the Pythagorean theorem to determine the magnitude of the //vector sum//. When finished, click the button to view the answer.

**Using Trigonometry to Determine a Vector's Direction** The direction of a //resultant// vector can often be determined by use of trigonometric functions. Most students recall the meaning of the useful mnemonic SOH CAH TOA from their course in trigonometry. SOH CAH TOA is a mnemonic that helps one remember the meaning of the three common trigonometric functions - sine, cosine, and tangent functions. These three functions relate an acute angle in a right triangle to the ratio of the lengths of two of the sides of the right triangle. The **sine function** relates the measure of an acute angle to the ratio of the length of the side opposite the angle to the length of the hypotenuse. The **cosine function** relates the measure of an acute angle to the ratio of the length of the side adjacent the angle to the length of the hypotenuse. The **tangent function** relates the measure of an angle to the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. The three equations below summarize these three functions in equation form. These three trigonometric functions can be applied to the [|hiker problem] in order to determine the direction of the hiker's overall displacement. The process begins by the selection of one of the two angles (other than the right angle) of the triangle. Once the angle is selected, any of the three functions can be used to find the measure of the angle. Write the function and proceed with the proper algebraic steps to solve for the measure of the angle. The work is shown below. Once the measure of the angle is determined, the direction of the vector can be found. In this case the vector makes an angle of 45 degrees with due East. Thus, the direction of this vector is written as 45 degrees. (Recall from [|earlier in this lesson] that the direction of a vector is the counterclockwise angle of rotation that the vector makes with due East.) The measure of an angle as determined through use of SOH CAH TOA is __not__ always the direction of the vector. The following vector addition diagram is an example of such a situation. Observe that the angle within the triangle is determined to be 26.6 degrees using SOH CAH TOA. This angle is the southward angle of rotation that the vector R makes with respect to West. Yet the direction of the vector as expressed with the CCW (counterclockwise from East) convention is 206.6 degrees.

Test your understanding of the use of SOH CAH TOA to determine the vector direction by trying the following two practice problems. In each case, use SOH CAH TOA to determine the direction of the resultant. When finished, click the button to view the answer.

In the above problems, the magnitude and direction of the sum of two vectors is determined using the Pythagorean theorem and trigonometric methods (SOH CAH TOA). The procedure is restricted to the addition of __two vectors that make right angles to each other__. When the two vectors that are to be added do not make right angles to one another, or when there are more than two vectors to add together, we will employ a method known as the head-to-tail vector addition method. This method is described below.

**Use of Scaled Vector Diagrams to Determine a Resultant** The magnitude and direction of the sum of two or more vectors can also be determined by use of an accurately drawn scaled vector diagram. Using a scaled diagram, the **head-to-tail method** is employed to determine the vector sum or resultant. A common Physics lab involves a //vector walk//. Either using centimeter-sized displacements upon a map or meter-sized displacements in a large open area, a student makes several consecutive displacements beginning from a designated starting position. Suppose that you were given a map of your local area and a set of 18 directions to follow. Starting at //home base//, these 18 displacement vectors could be //added together// in consecutive fashion to determine the result of adding the set of 18 directions. Perhaps the first vector is measured 5 cm, East. Where this measurement ended, the next measurement would begin. The process would be repeated for all 18 directions. Each time one measurement ended, the next measurement would begin. In essence, you would be using the head-to-tail method of vector addition.

The head-to-tail method involves [|drawing a vector to scale] on a sheet of paper beginning at a designated starting position. Where the head of this first vector ends, the tail of the second vector begins (thus, //head-to-tail// method). The process is repeated for all vectors that are being added. Once all the vectors have been added head-to-tail, the resultant is then drawn from the tail of the first vector to the head of the last vector; i.e., from start to finish. Once the resultant is drawn, its length can be measured and converted to //real// units using the given scale. The [|direction] of the resultant can be determined by using a protractor and measuring its counterclockwise angle of rotation from due East.

A step-by-step method for applying the head-to-tail method to determine the sum of two or more vectors is given below.
 * 1) Choose a scale and indicate it on a sheet of paper. The best choice of scale is one that will result in a diagram that is as large as possible, yet fits on the sheet of paper.
 * 2) Pick a starting location and draw the first vector //to scale// in the indicated direction. Label the magnitude and direction of the scale on the diagram (e.g., SCALE: 1 cm = 20 m).
 * 3) Starting from where the head of the first vector ends, draw the second vector //to scale// in the indicated direction. Label the magnitude and direction of this vector on the diagram.
 * 4) Repeat steps 2 and 3 for all vectors that are to be added
 * 5) Draw the resultant from the tail of the first vector to the head of the last vector. Label this vector as **Resultant** or simply **R**.
 * 6) Using a ruler, measure the length of the resultant and determine its magnitude by converting to real units using the scale (4.4 cm x 20 m/1 cm = 88 m).
 * 7) Measure the direction of the resultant using the counterclockwise convention discussed [|earlier in this lesson].

An example of the use of the head-to-tail method is illustrated below. The problem involves the addition of three vectors: 20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg.SCALE: 1 cm = 5 m The head-to-tail method is employed as described above and the resultant is determined (drawn in red). Its magnitude and direction is labeled on the diagram. SCALE: 1 cm = 5 m

Interestingly enough, the order in which three vectors are added has no affect upon either the magnitude or the direction of the resultant. The resultant will still have the same magnitude and direction. For example, consider the addition of the same three vectors in a different order.
 * 15 m, 210 deg. + 25 m, 300 deg. + 20 m, 45 deg.****SCALE: 1 cm = 5 m**

When added together in this different order, these same three vectors still produce a resultant with the same magnitude and direction as before (20. m, 312 degrees). The order in which vectors are added using the head-to-tail method is insignificant. SCALE: 1 cm = 5 m



**__10/13/11__**
__Lesson 1 C & D__ C) Resultants The **resultant** is the vector sum of two or more vectors. It is //the result// of adding two or more vectors together. If displacement vectors A, B, and C are added together, the result will be vector R. As shown in the diagram, vector R can be determined by the use of an [|accurately drawn, scaled, vector addition diagram]. To say that vector R is the //resultant displacement// of displacement vectors A, B, and C is to say that a person who walked with displacements A, then B, and then C would be displaced by the same amount as a person who walked with displacement R. Displacement vector R gives the same //result// as displacement vectors A + B + C. That is why it can be said that **A + B + C = R**

The above discussion pertains to the result of adding displacement vectors. When displacement vectors are added, the result is a //resultant displacement//. But any two vectors can be added as long as they are the same vector quantity. If two or more velocity vectors are added, then the result is a //resultant velocity//. If two or more force vectors are added, then the result is a //resultant force//. If two or more momentum vectors are added, then the result is ... In all such cases, the resultant vector (whether a displacement vector, force vector, velocity vector, etc.) is the result of adding the individual vectors. It is the same thing as adding A + B + C + ... . "To do A + B + C is the same as to do R." As an example, consider a football player who gets hit simultaneously by three players on the opposing team (players A, B, and C). The football player experiences three different applied forces. Each applied force contributes to a total or resulting force. If the three forces are added together using methods of vector addition ([|discussed earlier]), then the resultant vector R can be determined. In this case, to experience the three forces A, B and C is the same as experiencing force R. To be hit by players A, B, and C would result in the same force as being hit by one player applying force R. "To do A + B + C is the same as to do R." Vector R is the same result as vectors A + B + C!!

In summary, the resultant is the vector sum of all the individual vectors. The resultant is the result of combining the individual vectors together. The resultant can be determined by adding the individual forces together using [|vector addition methods].

D) Vector Components A [|vector] is a quantity that has both magnitude and direction. [|Displacement], [|velocity], [|acceleration], and [|force] are the vector quantities that we have discussed thus far in the Physics Classroom Tutorial. In the first couple of units, all vectors that we discussed were simply directed up, down, left or right. When there was a [|free-][|body][|diagram] depicting the forces acting upon an object, each individual force was directed in //one dimension// - either up or down or left or right. When an object had an acceleration and we described its direction, it was directed in //one dimension// - either up or down or left or right. Now in this unit, we begin to see examples of vectors that are directed in //two dimensions// - upward and rightward, northward and westward, eastward and southward, etc.

In situations in which vectors are directed at angles to the customary coordinate axes, a useful mathematical trick will be employed to //transform// the vector into two parts with each part being directed along the coordinate axes. For example, a vector that is directed northwest can be thought of as having two parts - a northward part and a westward part. A vector that is directed upward and rightward can be thought of as having two parts - an upward part and a rightward part. Any vector directed in two dimensions can be thought of as having an influence in two different directions. That is, it can be thought of as having two parts. Each part of a two-dimensional vector is known as a **component**. The components of a vector depict the influence of that vector in a given direction. The combined influence of the two components is equivalent to the influence of the single two-dimensional vector. The single two-dimensional vector could be replaced by the two components.

If Fido's dog chain is stretched upward and rightward and pulled tight by his master, then the tension force in the chain has two components - an upward component and a rightward component. To Fido, the influence of the chain on his body is equivalent to the influence of two chains on his body - one pulling upward and the other pulling rightward. If the single chain were replaced by two chains. with each chain having the magnitude and direction of the components, then Fido would not know the difference. This is not because Fido is //dumb// (a quick glance at his picture reveals that he is certainly not that), but rather because the combined influence of the two components is equivalent to the influence of the single two-dimensional vector.

Consider a picture that is hung to a wall by means of two wires that are stretched vertically and horizontally. Each wire exerts a tension force upon the picture to support its weight. Since each wire is stretched in two dimensions (both vertically and horizontally), the tension force of each wire has two components - a vertical component and a horizontal component. Focusing on the wire on the left, we could say that the wire has a leftward and an upward component. This is to say that the wire on the left could be replaced by two wires, one pulling leftward and the other pulling upward. If the single wire were replaced by two wires (each one having the magnitude and direction of the components), then there would be no affect upon the stability of the picture. The combined influence of the two components is equivalent to the influence of the single two-dimensional vector.



Consider an airplane that is flying from Chicago's O'Hare International Airport to a destination in Canada. Suppose that the plane is flying in such a manner that its resulting displacement vector is northwest. If this is the case, then the displacement of the plane has two components - a component in the northward direction and a component in the westward direction. This is to say that the plane would have the same displacement if it were to take the trip into Canada in two segments - one directed due North and the other directed due West. If the single displacement vector were replaced by these two individual displacement vectors, then the passengers in the plane would end up in the same final position. The combined influence of the two components is equivalent to the influence of the single two-dimensional displacement.

Any vector directed in two dimensions can be thought of as having two different components. The component of a single vector describes the influence of that vector in a given direction. In the [|next part of this lesson], we will investigate two methods for determining the magnitude of the components. That is, we will investigate //how much// influence a vector exerts in a given direction.

**__10/17/11__**
__Lesson 1 E__ E) Vector Resolution As mentioned [|earlier in this lesson], any vector directed at an angle to the horizontal (or the vertical) can be thought of as having two parts (or components). That is, any vector directed in two dimensions can be thought of as having two components. For example, if a chain pulls upward at an angle on the collar of a dog, then there is a tension force directed in two dimensions. This tension force has two components: an upward component and a rightward component. As another example, consider an airplane that is displaced northwest from O'Hare International Airport (in Chicago) to a destination in Canada. The displacement vector of the plane is in two dimensions (northwest). Thus, this displacement vector has two components: a northward component and a westward component.

In this unit, we learn two basic methods for determining the magnitudes of the components of a vector directed in two dimensions. The process of determining the magnitude of a vector is known as **vector resolution**. The two methods of vector resolution that we will examine are
 * the parallelogram method
 * [|the trigonometric method]

**Parallelogram Method of Vector Resolution** The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector. Briefly put, the method involves drawing the vector to scale in the indicated direction, sketching a parallelogram around the vector such that the vector is the diagonal of the parallelogram, and determining the magnitude of the components (the sides of the parallelogram) using the scale. If one desires to determine the components as directed along the traditional x- and y-coordinate axes, then the parallelogram is a rectangle with sides that stretch vertically and horizontally. A step-by-step procedure for using the parallelogram method of vector resolution is: The step-by-step procedure above is illustrated in the diagram below to show how a velocity vector with a magnitude of 50 m/s and a direction of 60 degrees above the horizontal may be resolved into two components. The diagram shows that the vector is first [|drawn to scale] in the indicated direction; a parallelogram is sketched about the vector; the components are labeled on the diagram; and the result of measuring the length of the vector components and converting to m/s using the scale. (NOTE: because different computer monitors have different resolutions, the actual length of the vector on your monitor may not be 5 cm.)
 * 1) Select a scale and accurately draw the vector to scale in the indicated direction.
 * 2) Sketch a parallelogram around the vector: beginning at the [|tail] of the vector, sketch vertical and horizontal lines; then sketch horizontal and vertical lines at the [|head] of the vector; the sketched lines will meet to form a rectangle (a special case of a parallelogram).
 * 3) Draw the components of the vector. The components are the //sides// of the parallelogram. The tail of the components start at the tail of the vector and stretches along the axes to the nearest corner of the parallelogram. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward velocity component might be labeled vx; etc.
 * 5) Measure the length of the sides of the parallelogram and [|use the scale to determine the magnitude] of the components in //real// units. Label the magnitude on the diagram.

**Trigonometric Method of Vector Resolution** The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector. [|Earlier in lesson 1], the use of trigonometric functions to determine the direction of a vector was described. Now in this part of lesson 1, trigonometric functions will be used to determine the components of a single vector. [|Recall from the earlier discussion] that trigonometric functions relate the ratio of the lengths of the sides of a right triangle to the measure of an acute angle within the right triangle. As such, trigonometric functions can be used to determine the length of the sides of a right triangle if an angle measure and the length of one side are known. The method of employing trigonometric functions to determine the components of a vector are as follows: The above method is illustrated below for determining the components of the force acting upon Fido. As the 60-Newton tension force acts upward and rightward on Fido at an angle of 40 degrees, the components of this force can be determined using trigonometric functions.
 * 1) Construct a //rough// sketch (no scale needed) of the vector in the indicated direction. Label its magnitude and the angle that it makes with the horizontal.
 * 2) Draw a rectangle about the vector such that the vector is the diagonal of the rectangle. Beginning at the [|tail] of the vector, sketch vertical and horizontal lines. Then sketch horizontal and vertical lines at the [|head] of the vector. The sketched lines will meet to form a rectangle.
 * 3) Draw the components of the vector. The components are the //sides// of the rectangle. The tail of each component begins at the tail of the vector and stretches along the axes to the nearest corner of the rectangle. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward force velocity component might be labeled vx; etc.
 * 5) To determine the length of the side opposite the indicated angle, use the sine function. Substitute the magnitude of the vector for the length of the hypotenuse. Use some algebra to solve the equation for the length of the side opposite the indicated angle.
 * 6) Repeat the above step using the cosine function to determine the length of the side adjacent to the indicated angle.

In conclusion, a vector directed in two dimensions has two components - that is, an influence in two separate directions. The amount of influence in a given direction can be determined using methods of vector resolution. Two methods of vector resolution have been described here - [|a graphical method] (parallelogram method) and a [|trigonometric method].

__**10/18/11**__
__Lesson 1 G & H__ G) Relative Velocity and Riverboat Problems On occasion objects move within a medium that is moving with respect to an observer. For example, an airplane usually encounters a wind - air that is moving with respect to an observer on the ground below. As another example, a motorboat in a river is moving amidst a river current - water that is moving with respect to an observer on dry land. In such instances as this, the magnitude of the velocity of the moving object (whether it be a plane or a motorboat) with respect to the observer on land will not be the same as the speedometer reading of the vehicle. That is to say, the speedometer on the motorboat might read 20 mi/hr; yet the motorboat might be moving relative to the observer on shore at a speed of 25 mi/hr. Motion is relative to the observer. The observer on land, often named (or misnamed) the "stationary observer" would measure the speed to be different than that of the person on the boat. The observed speed of the boat must always be described relative to who the observer is. To illustrate this principle, consider a plane flying amidst a **tailwind**. A tailwind is merely a wind that approaches the plane from behind, thus increasing its resulting velocity. If the plane is traveling at a velocity of 100 km/hr with respect to the air, and if the wind velocity is 25 km/hr, then what is the velocity of the plane relative to an observer on the ground below? The resultant velocity of the plane (that is, the result of the wind velocity contributing to the velocity due to the plane's motor) is the vector sum of the velocity of the plane and the velocity of the wind. This resultant velocity is quite easily determined if the wind approaches the plane directly from behind. As shown in the diagram below, the plane travels with a resulting velocity of 125 km/hr relative to the ground. If the plane encounters a headwind, the resulting velocity will be less than 100 km/hr. Since a headwind is a wind that approaches the plane from the front, such a wind would decrease the plane's resulting velocity. Suppose a plane traveling with a velocity of 100 km/hr with respect to the air meets a headwind with a velocity of 25 km/hr. In this case, the resultant velocity would be 75 km/hr; this is the velocity of the plane relative to an observer on the ground. This is depicted in the diagram below. Now consider a plane traveling with a velocity of 100 km/hr, South that encounters a **side wind** of 25 km/hr, West. Now what would the resulting velocity of the plane be? This question can be answered in the same manner as the previous questions. The resulting velocity of the plane is the vector sum of the two individual velocities. To determine the resultant velocity, the plane velocity (relative to the air) must be added to the wind velocity. This is the same procedure that was used above for the headwind and the tailwind situations; only now, the resultant is not as easily computed. Since the two vectors to be added - the southward plane velocity and the westward wind velocity - are at right angles to each other, the [|Pythagorean theorem] can be used. This is illustrated in the diagram below. In this situation of a side wind, the southward vector can be added to the westward vector using the [|usual methods of vector addition]. The magnitude of the resultant velocity is determined using Pythagorean theorem. The algebraic steps are as follows: (100 km/hr)2 + (25 km/hr)2 = R2 10 000 km2/hr2 + 625 km2/hr2 = R2  10 625 km2/hr2 = R2  SQRT(10 625 km2/hr2) = R  **103.1 km/hr = R**

The direction of the resulting velocity can be determined using a [|trigonometric function]. Since the plane velocity and the wind velocity form a right triangle when added together in head-to-tail fashion, the angle between the resultant vector and the southward vector can be determined using the sine, cosine, or tangent functions. The tangent function can be used; this is shown below:

tan (theta) = (opposite/adjacent) tan (theta) = (25/100) theta = invtan (25/100) **theta = 14.0 degrees** If the resultant velocity of the plane makes a 14.0 degree angle with the southward direction (theta in the above diagram), then the direction of the resultant is 256 degrees. Like any vector, the resultant's [|direction] is measured as a counterclockwise angle of rotation from due East.

**Analysis of a Riverboat's Motion** The affect of the wind upon the plane is similar to the affect of the river current upon the motorboat. If a motorboat were to head straight across a river (that is, if the boat were to point its bow straight towards the other side), it would not reach the shore directly across from its starting point. The river current influences the motion of the boat and carries it downstream. The motorboat may be moving with a velocity of 4 m/s directly across the river, yet the resultant velocity of the boat will be greater than 4 m/s and at an angle in the downstream direction. While the speedometer of the boat may read 4 m/s, its speed with respect to an observer on the shore will be greater than 4 m/s. The resultant velocity of the motorboat can be determined in the same manner as was done for the plane. The resultant velocity of the boat is the vector sum of the boat velocity and the river velocity. Since the boat heads straight across the river and since the current is always directed straight downstream, the two vectors are at right angles to each other. Thus, the [|Pythagorean theorem] can be used to determine the resultant velocity. Suppose that the river was moving with a velocity of 3 m/s, North and the motorboat was moving with a velocity of 4 m/s, East. What would be the resultant velocity of the motorboat (i.e., the velocity relative to an observer on the shore)? The magnitude of the resultant can be found as follows: (4.0 m/s)2 + (3.0 m/s)2 = R2 16 m2/s2 + 9 m2/s2 = R2  25 m2/s2 = R2  SQRT (25 m2/s2) = R  **5.0 m/s = R**

The [|direction] of the resultant is the counterclockwise angle of rotation that the resultant vector makes with due East. This angle can be determined using a trigonometric function as shown below. tan (theta) = (opposite/adjacent) tan (theta) = (3/4) theta = invtan (3/4) **theta = 36.9 degrees**

Given a boat velocity of 4 m/s, East and a river velocity of 3 m/s, North, the resultant velocity of the boat will be 5 m/s at 36.9 degrees.

Motorboat problems such as these are typically accompanied by three separate questions: The first of these three questions was answered above; the resultant velocity of the boat can be determined using the Pythagorean theorem (magnitude) and a trigonometric function (direction). The second and third of these questions can be answered using the [|average speed equation] (and a lot of logic). **ave. speed = distance/time** Consider the following example.  The solution to the first question has already been shown in the [|above discussion]. The resultant velocity of the boat is 5 m/s at 36.9 degrees. We will start in on the second question. The river is 80-meters wide. That is, the distance from shore to shore as measured straight across the river is 80 meters. The time to cross this 80-meter wide river can be determined by rearranging and substituting into the average speed equation. **time = distance /(ave. speed)** The distance of 80 m can be substituted into the numerator. But what about the denominator? What value should be used for average speed? Should 3 m/s (the current velocity), 4 m/s (the boat velocity), or 5 m/s (the resultant velocity) be used as the average speed value for covering the 80 meters? With what average speed is the boat traversing the 80 meter wide river? Most students want to use the resultant velocity in the equation since that is the actual velocity of the boat with respect to the shore. Yet the value of 5 m/s is the speed at which the boat covers the diagonal dimension of the river. And the diagonal distance across the river is not known in this case. If one knew the **distance C** in the diagram below, then the **average speed C** could be used to calculate the time to reach the opposite shore. Similarly, if one knew the **distance B** in the diagram below, then the **average speed B** could be used to calculate the time to reach the opposite shore. And finally, if one knew the **distance A** in the diagram below, then the **average speed A** could be used to calculate the time to reach the opposite shore. In our problem, the 80 m corresponds to the distance A, and so the average speed of 4 m/s (average speed in the direction straight across the river) should be substituted into the equation to determine the time. **time = (80 m)/(4 m/s) = 20 s** It requires 20 s for the boat to travel across the river. During this 20 s of crossing the river, the boat also drifts downstream. [|Part c of the problem] asks "What distance downstream does the boat reach the opposite shore?" The same equation must be used to calculate this //downstream distance//. And once more, the question arises, which one of the three average speed values must be used in the equation to calculate the distance downstream? The distance downstream corresponds to **Distance B** on the above diagram. The speed at which the boat covers this distance corresponds to **Average Speed B** on the diagram above (i.e., the speed at which the current moves - 3 m/s). And so the average speed of 3 m/s (average speed in the downstream direction) should be substituted into the equation to determine the distance. **distance = ave. speed * time = (3 m/s) * (20 s)** **distance = 60 m** The boat is carried 60 meters downstream during the 20 seconds it takes to cross the river.
 * 1) What is the resultant velocity (both magnitude and direction) of the boat?
 * 2) If the width of the river is //X// meters wide, then how much time does it take the boat to travel shore to shore?
 * 3) What distance downstream does the boat reach the opposite shore?
 * Example 1 A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North.
 * 1) What is the resultant velocity of the motorboat?
 * 2) If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?
 * 3) What distance downstream does the boat reach the opposite shore? ||

The mathematics of the above problem is no more difficult than dividing or multiplying two numerical quantities by each other. The mathematics is easy! The difficulty of the problem is conceptual in nature; the difficulty lies in deciding which numbers to use in the equations. That decision emerges from one's conceptual understanding (or unfortunately, one's misunderstanding) of the complex motion that is occurring. The motion of the riverboat can be divided into two simultaneous parts - a motion in the direction straight across the river and a motion in the downstream direction. These two parts (or components) of the motion occur simultaneously for the same time duration (which was 20 seconds in the above problem). The decision as to which velocity value or distance value to use in the equation must be consistent with the [|diagram above]. The boat's motor is what carries the boat across the river the **Distance A** ; and so any calculation involving the **Distance A** must involve the speed value labeled as **Speed A** (the boat speed relative to the water). Similarly, it is the current of the river that carries the boat downstream for the **Distance B** ; and so any calculation involving the **Distance B** must involve the speed value labeled as **Speed B** (the river speed). Together, these two parts (or components) add up to give the resulting motion of the boat. That is, the across-the-river component of displacement adds to the downstream displacement to equal the resulting displacement. And likewise, the boat velocity (across the river) adds to the river velocity (down the river) to equal the resulting velocity. And so any calculation of the Distance C or the Average Speed C ("Resultant Velocity") can be performed using the Pythagorean theorem.

Now to illustra te an important point, let's try a second example problem that is similar to the [|first example problem]. Make an attempt to answer the three questions and then click the button to c heck your answer.
 * Example 2 A motorboat traveling 4 m/s, East encounters a current traveling 7.0 m/s, North.
 * 1) What is the resultant velocity of the motorboat?
 * 2) If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?
 * 3) What distance downstream does the boat reach the opposite shore? ||

 An import ant concept emerges from the analysis of the two example problems above. In Example 1, the time to cross the 80-meter wide river (when moving 4 m/s) was 20 seconds. This was in the presence of a 3 m/s current velocity. In Example 2, the current velocity was much greater - 7 m/s - yet the time to cross the river remained unchanged. In fact, the current velocity itself has no affect upon the time required for a boat to cross the river. The river moves downstream parallel to the banks of the river. As such, there is no way that the current is capable of assisting a boat in crossing a river. While the increased current may affect the resultant velocity - making the boat travel with a greater speed with respect to an observer on the ground - it does not increase the speed in the direction across the river. The component of the resultant velocity that is increased is the component that is in a direction pointing down the river. It is often said that "perpendicular components of motion are independent of each other." As applied to riverboat problems, this would mean that an across-the-river variable would be independent of (i.e., not be affected by) a downstream variable. The time to cross the river is dependent upon the velocity at which the boat crosses the river. It is only the component of motion directed across the river (i.e., the boat velocity) that affects the time to travel the distance directly across the river (80 m in this case). The component of motion perpendicular to this direction - the current velocity - only affects the distance that the boat travels down the river. This concept of perpendicular components of motion will be investigated in more detail in the [|next part of Lesson 1].

H) Independence of Perpendicular Components of Motion A force vector that is directed upward and rightward has two parts - an upward part and a rightward part. That is to say, if you pull upon an object in an upward and rightward direction, then you are exerting an influence upon the object in two separate directions - an upward direction and a rightward direction. These two parts of the two-dimensional vector are referred to as [|components] . A **component** describes the affect of a single vector in a given direction. Any force vector that is exerted at an angle to the horizontal can be considered as having two parts or components. The vector sum of these two components is always equal to the force at the given angle. This is depicted in the diagram below.

Any vector - whether it is a force vector, displacement vector, velocity vector, etc. - directed at an angle can be thought of as being composed of two perpendicular components. These two components can be represented as legs of a right triangle formed by projecting the vector onto the x- and y-axis. The two perpendicular parts or components of a vector are independent of each other. Consider the pull upon Fido as an example. If the horizontal pull upon Fido increases, then Fido would be accelerated at a greater rate to the right; yet this greater horizontal pull would not exert any vertical influence upon Fido. Pulling horizontally with more force does not lift Fido vertically off the ground. A change in the horizontal component does not affect the vertical component. This is what is meant by the phrase "perpendicular components of vectors are independent of each other." A change in one component does not affect the other component. Changing a component will affect the motion in that specific direction. While the change in one of the components will alter the magnitude of the resulting force, it does not alter the magnitude of the other component. The resulting motion of a plane flying in the presence of a crosswind is the combination (or sum) of two simultaneous velocity vectors that are perpendicular to each other. Suppose that a plane is attempting to fly northward from Chicago to the Canada border by simply directing the plane due northward. If the plane encounters a crosswind directed towards the west, then the resulting velocity of the plane would be northwest. The northwest velocity vector consists of two components - a north component resulting from the plane's motor (the //plane velocity//) and a westward component resulting from the crosswind (the //wind velocity//). These two components are independent of each other. An alteration in one of the components will not affect the other component. For instance, if the wind velocity increased, then the plane would still be covering ground in the northerly direction at the same rate. It is true that the alteration of the wind velocity would cause the plane to travel more westward; however, the plane still flies northward at the same speed. Perpendicular components of motion do not affect each other. Now consider an air balloon descending through the air toward the ground in the presence of a wind that blows eastward. Suppose that the downward velocity of the balloon is 3 m/s and that the wind is blowing east with a velocity of 4 m/s. The resulting velocity of the air balloon would be the combination (i.e., the vector sum) of these two simultaneous and independent velocity vectors. The air balloon would be moving downward and eastward. If the wind velocity increased, the air balloon would begin moving faster in the eastward direction, but its downward velocity would not be altered. If the balloon were located 60 meters above the ground and was moving downward at 3 m/s, then it would take a time of 20 seconds to travel this vertical distance. **d = v • t** So **t = d / v** (60 m) / (3 m/s) **20 seconds** During the 20 seconds taken by the air balloon to fall the 60 meters to the ground, the wind would be carrying the balloon in the eastward direction. With a wind speed of 4 m/s, the distance traveled eastward in 20 seconds would be 80 meters. If the wind speed increased from the value of 4 m/s to a value of 6 m/s, then it would still take 20 seconds for the balloon to fall the 60 meters of downward distance. A motion in the downward direction is affected only by downward components of motion. An alteration in a horizontal component of motion (such as the eastward wind velocity) will have no affect upon vertical motion. Perpendicular components of motion are independent of each other. A variation of the eastward wind speed from a value of 4 m/s to a value of 6 m/s would only cause the balloon to be blown eastward a distance of 120 meters instead of the original 80 meters.

In the [|most recent section of Lesson 1], the topic of relative velocity and riverboat motion was discussed. A boat on a river often heads straight across the river, perpendicular to its banks. Yet because of the flow of water (i.e., the current) moving parallel to the river banks, the boat does not land on the bank directly across from the starting location. The resulting motion of the boat is the combination (i.e., the vector sum) of these two simultaneous and independent velocity vectors - the boat velocity plus the river velocity. In the diagram at the right, the boat is depicted as moving eastward across the river while the river flows southward. The boat starts at Point A and heads itself towards Point B. But because of the flow of the river southward, the boat reaches the opposite bank of the river at Point C. The time required for the boat to cross the river from one side to the other side is dependent upon the boat velocity and the width of the river. Only an eastward component of motion could affect the time to move eastward across a river. Suppose that the boat velocity is 4 m/s and the river velocity is 3 m/s. The magnitude of the resultant velocity could be determined to be 5 m/s using the Pythagorean Theorem. The time required for the boat to cross a 60-meter wide river would be dependent upon the boat velocity of 4 m/s. It would require 15 seconds to cross the 60-meter wide river. **d = v • t** So **t = d / v** (60 m) / (4 m/s) **15 seconds** The southward river velocity will not affect the time required for the boat to travel in the eastward direction. If the current increased such that the river velocity became 5 m/s, then it would still require 15 seconds to cross the river. Perpendicular components of motion are independent of each other. An increase in the river velocity would simply cause the boat to travel further in the southward direction during these 15 seconds of motion. An alteration in a southward component of motion only affects the southward motion.

All vectors can be thought of as having perpendicular components. In fact, any motion that is at an angle to the horizontal or the vertical can be thought of as having two perpendicular motions occurring simultaneously. These perpendicular components of motion occur independently of each other. Any component of motion occurring strictly in the horizontal direction will have no affect upon the motion in the vertical direction. Any alteration in one set of these components will have no affect on the other set. In [|Lesson 2], we will apply this principle to the motion of projectiles that typically encounter both horizontal and vertical motion.

__**10/19/11**__
__Lesson 2 A & B__ A) What is a Projectile?
 * P.)** A projectile is an object that is only affected by the forces of gravity. This reading will most likely be going into further detail about a projectile.

2. What kind of acceleration does gravity cause? 3. Why does a projectile travel a parabolic trajectory?
 * Q.)** 1. What does inertia have to do with projectiles?


 * S.)** This passage basically talks about how a projectile is any object in which the only force acting upon it is gravity. Without gravity, a projectile would continue to travel in the direction in which it was fired. Also, the passage talks about how a force is merely anything that keeps a constant acceleration, and does not need to keep the object the force is acting upon in motion.

2. Gravity is a force with a negative and constant acceleration. 3. A projectile travels in a parabolic trajectory because of the influence gravity has on the trajectory of the projectile. Since gravity causes a constant deceleration, the projectile in question would slow down and drop in altitude; the drop in height would be gradual at first, but as time passes, the drop would be more significant, and since gravity has a constant, negative acceleration, the trajectory becomes parabolic.
 * T.)** 1. Without gravity, an object will continue the direction it was fired in without stopping. The object would then be inert.

B) Characteristics of a Projectile's Trajectory
 * P.)** I predict that the passage will most likely be depicting how a projectile moves and why it moves in such a manner.

2. Does changing the angle in which a projectile is fired do anything to the trajectory? 3. Is there an angle that is proven to be the angle that allows for the furthest range possible?
 * Q.)** 1. What forces affect the trajectory of a projectile?


 * S.)** The main topic of this passage is the different kinds of launches for a projectile. There are three main types: launches from the same plane, launches from a higher plane to a lower plane, and launches from a higher plane to a lower plane but from an angle.

2. Changing the angle in which a projectile is fired at does affect the trajectory. An angle in between 0º and 90º allows for a further range of the projectile than a horizontal launch. 3. 45º is the angle in which a projectile will travel furthest.
 * T.)** 1. Gravity is the only force that acts upon a projectile.

**__10/20/11__**
__Lesson 2 C__ C) Horizontal and Vertical Components of Velocity and Displacement
 * P.)** The main ideas of this reading will be about how you describe projectile motion through numbers. We will learn about how the x and y parts of the velocity and displacement change within time or remain constant.

2. What is the difference between firing a projectile straight ahead (180º angle) or firing a projectile at an angle upwards (from 90º-180º)? 3. What is the difference between horizontal and vertical components?
 * Q.)** 1. How would understanding the way a projectile moves help with the understanding of describing projectile motion with numbers?


 * S.)** The two key principles in projectile motion are that the horizontal velocity is constant and that the vertical velocity is always -9.8 m/s. Also, the route a projectile will travel whenever fired at an upwards angle is always parabolic, so the vertical velocity of the projectile 1 second before reaching the peak point is the same as 1 second after reaching the peak. The vertical displacement in this motion can be predicted with the equation y=0.5gt 2 and g is always -9.8 m/s 2 and t is always time. In projectile motion, the horizontal movement is only influenced by the speed it was moving horizontally and the amount of time elapsed.

2. When you fire a projectile straight ahead, you fire from the max height. When you fire at an angle upwards (90º-180º), you will get a longer range as you have not yet reached max height and the initial velocity is the same as the one firing straight ahead. 3. The horizontal component is the final displacement of the projectile horizontally. The vertical component is the final displacement of the projectile vertically.
 * T.)** 1. When you understand how a projectile moves, you can visualize a diagram of the movement of the projectile in your head. The diagram in your head can then help you understand the numbers and what they represent.