Ch3_SekiK

=Physics Classroom Notes= toc

Lesson 1(ab):
Topic Sentence: Vectors, quantities with magnitude or direction, are represented on a diagram when they have a clear tail and head, when they have a scale, and when the direction is listed. A study of motion will involve the introduction of a variety of quantities that are used to describe the physical world. 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. Each of these vector quantities are unique in that a full description of the quantity demands that both a magnitude and a direction are listed. 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. 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. **Conventions for Describing Directions of Vectors** Vectors can be directed due East, due West, due South, and due North. 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.
 * A: Vectors and Direction**
 * Vectors and Direction**
 * 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).
 * 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.

**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. 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.

Topic Sentence: There are three ways to add vectors: through the pythagorean theorem, through trigonometry, and through the head to tail method. 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). Observe the following summations of two force vectors:
 * B: Vector Addition**
 * Vector Addition**

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. There are a variety of methods for determining the magnitude and direction of the result. The two methods that will be discussed in this lesson and used throughout the entire unit are: **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 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. The Pythagorean theorem can be used to determine the resultant. 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 **Using Trigonometry to Determine a Vector's Direction** The direction of a //resultant// vector can often be determined by use of trigonometric functions. These three functions, sine, cosine, and tangent, 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. 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. 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. Yet the direction of the vector as expressed with the CCW (counterclockwise from East) convention is 206.6 degrees.
 * the Pythagorean theorem and trigonometric methods
 * the head-to-tail method using a scaled vector diagram

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. **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.

A step-by-step method for applying the head-to-tail method to determine the sum of two or more vectors is given below. 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.
 * 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

Lesson 1(cd):
C: Resultants Topic Sentence: The Resultant is the sum of individual vectors and can be determined by adding the individual forces together using vector addition. The **resultant** is the vector sum of two or more vectors. It is //the result// of adding two or more vectors together. Vector R can be determined by the use of an accurately drawn, scaled, vector addition diagram. 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."

D: Vector Compenents Topic Sentence: Vectors that are have two directions are equivalent to two different vector components in the same two directions. 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. Any vector directed in two dimensions can be thought of as having an influence in two different directions. . 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. 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.

Lesson 1(e):
E: Vector Resolution Topic Sentence: There are two ways to find vector resolution: the parallelogram method, and the trigonometric method. Any vector directed in two dimensions can be thought of as having two components. 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. 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.
 * 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.
 * 3) Draw the components of the vector. The components are the //sides// of the parallelogram.
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side.
 * 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. Trigonometric functions will be used to determine the components of a single vector. 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:
 * 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. 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.
 * 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.

Lesson 1 (gh):
Topic Sentence: Velocity is different depending on the point of view, and when solving riverboat problems, use the correct velocity for the different distances. On occasion objects move within a medium that is moving with respect to an observer. 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
 * G: Relative Velocity and Riverboat Problems**

**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.

 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).
 * 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?

Consider the following example.  We will start in on the second question. The river is 80-meters wide. The time to cross this 80-meter wide river can be determined by rearranging and substituting into the average speed equation. 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. If one knew the distance C in t he diagram below, then the average speed C could be used to calculate the time to reach the opposite shore. 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 que stion 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. 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
 * 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? ||

Topic Sentence: Perpendicular components of Motion are independent of each other - a change in one doesn't affect the other. 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. A change in the horizontal component does not affect the vertical component.
 * H: Independence of Perpendicular Components of Motion**

Lesson 2 (ab)
A: What is projectile motion? Central Theme: Gravity is the only force acting upon a projectile, and it affects only the vertical motion of the object.
 * 1) Like the title says, what constitutes a projectiles?
 * 2) Are projectiles the same thing as free-falling objects?
 * 3) How are projectiles different from free-falling objects?
 * 4) What does inertia have to do with projectile motion?
 * 5) Does gravity change the path of a projectile significantly?
 * 1) A projectile is an object that is only acted upon by gravity
 * 2) According to each of their definitions, projectiles and free-falling objects are the same.
 * 3) They are not different
 * 4) Inertia allows an object to stay at a certain speed as long as there is nothing acting upon it. Thus, even in the presence of gravity, a projectile will still have the same speed as one not in gravity.
 * 5) Gravity does change the path of a projectile by making it fall, though it does nothing to slow it down.

B: Characteristics of a Projectile's trajectory Central Theme: Each component is independent of each other, which means that gravity will not affect the horizontal motion.
 * 1) What defines a projectile's trajectory?
 * 2) How does it differ from other things?
 * 3) Are the paths parabolic, like free falling objects?
 * 4) Do they have similar characteristics as free falling objects? (t1,v1,y1, t2,v2,y2....)
 * 5) How does gravity affect the path of a projectile?
 * 1) Projectiles move horizontally and vertically at the same time, though these two are independent of each other. The path is parabolic.
 * 2) It differs because the only force acting upon it is gravity.
 * 3) Yes, the path is parabolic.
 * 4) Yes, it does have these same characteristics.
 * 5) Gravity affects the path of a projectile by making it parabolic, making it fall. However, it will stay at the same horizontal speed.

Lesson 2 (c):
Central Theme: Because vector components are independent of each other, gravity will not affect the horizontal path. This means that the horizontal velocity will stay the same while it will accelerate downwards due to gravity.
 * 1) How do you calculate the horizontal and vertical velocities of a projectile?
 * 2) Is there a way to calculate the exact path of a projectile?
 * 3) What are the equations, if necessary, to find these?
 * 4) How are they similar to other equations we have used in the past?
 * 5) How does the sign of the upward/downward velocity change as the projectile goes in its path?
 * 1) The horizontal velocity will not change, so it is constant throughout. The vertical velocity can be found by using an equation.
 * 2) Not that we have learned yet. However, we can make rough estimates.
 * 3) The three equations we need to know are useful for displacement. For vertical displacement, the equation is y = .5gt^2. Horizontal displacement is x = vit. To find the vertical displacement for an angled projectile, the equation is y=(viy)(t) + .5gt^2.
 * 4) The vertical displacement equation is similar to d = vit + 1/2at^2. Because vi is 0 when it is essentially dropped in projectile problems, this makes sense. The x=vit equation is similar to a rewritten v = d/t equation. The vertical displacement for an angled projectile is exactly the same as d = vit +1/2at^2.
 * 5) For an angled projectile, the acceleration will change from negative to positive as it goes towards its max height and then away.

=Class Notes=

10/14
Collinear Velocities are just added or subtracted depending on if they're going opposite directions or not. Resultant - answer to a vector addition problem To get resultant (head to tail method) Tail to tail method (estimation)
 * 1) Draw a sketch of the vector __head to tail__.
 * 2) Draw the resultant from tail of first to end.
 * 3) Do pythagorean theorem to find resultant side
 * 4) Do trig to find the angle
 * 1) Size must be between minimum and maximum of two vectors.
 * 2) Angles must be between the angles of vectors given.

Lab: Vector Addition
Analysis: == Our percent error was very very high for this experiment. We very likely made a mistake somewhere in our measurements, as that is the only conceivable way to get a percent error of over 10%, much less 22.6%. What probably caused this was the fact that we did not start from the origin when measuring our resultant. We started on one side of the pillar, but measured from the opposite one. This threw off our results by an insane amount. We could have fixed our results by thinking about our resultant more carefully and understanding that it is only applicable when measured from the origin. I would hate to be the group who got our card.

LAB: Vector Addition (continued)
Our percent error shows that we were very accurate in our procedure. As opposed to the last time, in which we were off by quite a bit, we had less than 2% error on this trial. This is likely because we took greater in getting accurate measurement. However, we could have made this more accurate. Because the resultant length was very long, it was hard to line it up exactly to the origin. We could have avoided this slight error by having someone stand at the origin to make sure that we were lined up exactly.

Ball in Cup Activity
We were slightly off in our cup placement. Every time we launched our ball, it would hit the bottom of the cup, but never go inside. We likely messed up somewhere in our calculations to create this rift. Either that, or we didn't measure accurately enough to get good results. We could have fixed this problem by being more meticulous in both our measurements and calculations. There really is no excuse for this and no other cure other than to be more careful next time. video?

LAB: Shoot your Grade
Rationale w/ Purpose: What we are aiming to do is to launch the ball in such a way that it goes through five hoops and lands in a cup. Our given angle was 25 degrees. Because the launcher is set on top of the cabinets, this would be an off the cliff problem, which we have already covered in class.

Hypothesis: If we do all of our calculations, then the ball should smoothly sail between each hoop and land in the cup at the end.

Materials: The materials that we received were a launcher, cup, plunger, ball, rolls of tape, tape, measuring tape, string, paper, and carbon paper.

Method: Using the preset launcher at the given angle of 25 degrees, we calculated the initial velocity by setting up the equation, finding vit, and then plugging that into d = vit + 1/2at^2 equation to find the total time. We then took the number we had for vit and plugged in the time we got in order to get the initial velocity. To make sure that our ball went into the cup and not hit it, we subtracted the height of the cup from the total y distance. This gave us the distance where the ball would land inside. The next task we had to do was to calculate where we had to place the rings. The group preceding us had already set up rings at 50, 100, 151, 208, and 258 centimeters from the launcher. This fit in with our total distance of 313 centimeters perfectly, as each was about 50 centimeters apart. Using this distance as well as the velocity we had calculated before, we were able to find what the y displacement was. This enabled us to put the rings exactly where we wanted them. Once we had them in place, it was a matter of tweaking the rings to the launcher's inconsistencies.

Observations: Our average distance was 3.21 m, or 321 cm. Our launcher is relatively consistent, though it is still a little bit off sometimes. We will use this total distance to calculate initial velocity as well as the placement of the cup.
 * Trials || Distance (m) ||
 * Trial 1 || 3.25 ||
 * Trial 2 || 3.20 ||
 * Trial 3 || 3.22 ||
 * Trial 4 || 3.17 ||
 * Trial 5 || 3.22 ||

Calculations:

Video: media type="file" key="Shoot your grade lab video.m4v" width="300" height="300"
 * Rings/Cup || Did it make it through? ||
 * Ring #1 || Yes ||
 * Ring #2 || Yes ||
 * Ring #3 || Yes ||
 * Ring #4 || Yes ||
 * Ring #5 || Yes ||
 * Cup || YES ||


 * || X (cm) || Time (s) || Theoretical Y in cm || Experimental Y in cm || % error ||
 * Ring #1 || 50 || .1147 || 16.9 || 21.5 || 27.2% ||
 * Ring #2 || 100 || .2294 || 20.85 || 24.0 || 15.1% ||
 * Ring #3 || 151 || .346 || 11.65 || - || - ||
 * Ring #4 || 208 || .477 || -14.6 || -15.5 || 6.16% ||
 * Ring #5 || 258 || .592 || -51.73 || -56.0 || 8.25% ||
 * Cup || 313 || .737 || 0 || 0 || 0% ||

% error: Though we made it through each ring and landed in the cup, it is clear that our percent error is rather high. This is because the screw making sure that the angle was steady was broken; so each time we loaded the ball in, the angle increased slightly. This created the much higher experimental values that you can see here.

Conclusion: Our ball was able to go through each ring to eventually land in the cup. It followed a parabolic path as we thought it would. This is very clear from the ring placements. There is a clear max height, and it steeply drops afterwards to make a parabolic curve. Knowing this, we were able to calculate the placement of the rings and the cup so that it would sail through. However, there are clearly errors throughout our ring placement. As I already discussed, loading the launcher changed our angle, causing the large percent error. There is also the fact that the launcher is inconsistent in its launch. Another source of relatively small error is the fact that all of our calculations are based on the assumption of no other force other than gravity. However, in the real world, there are other forces, such as air resistance. This could have played a part in making this lab more difficult. To fix this, we can do many things. To secure the angle, the easiest way to fix it would be to replace the broken screw, making the angle more consistent in between each launch. This would have allowed us to complete the lab faster and more accurately. To fix the inconsistency in the launcher, the best way to alleviate that problem would be to conduct multiple trials. If you do more trials, then the average will eventually win out over the outliers. To have a completely accurate launch, however, we have to look at the problem of air resistance. Though it is a minimal factor, it does affect the ball when it shouldn't due to our previous calculations. The only way to fix this problem would be to conduct the test in a vacuum, though this is highly unrealistic. This is a very important concept to understand if you are a football quarterback. As a quarterback, you have to throw the ball in an accurate, consistent way to avoid many obstacles as well as getting the ball to your target. If the ball hits something on its path, it will never reach the intended target. Thus, one has to throw in such a way to sail through the arms of the opponent and into the hands of your wide receiver.