Ch3_KwarkR

toc Chapter 3

Reading Notes: Vectors Lesson 1 (a,b)
a)Vectors and Direction A study of motion involves quantities that are used to describe the physical world. Examples include distance, displacement, speed, velocity, acceleration, force, mass, momentum, energy, work, power, etc. Vector quantities are often represented by scaled __ [|vector diagrams] __. They are also called free-body diagrams, and depict a vector by use of an arrow drawn to scale in a specific direction. The vector diagram depicts a displacement vector. 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** .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: **Representing the Magnitude of a Vector** The magnitude of a vector in a scaled vector diagram is depicted by the length of the arrow, which is drawn a precise length in accordance with a chosen scale.
 * 1) The direction of a vector is expressed as an angle of rotation of the vector about its "__ [|tail] __" from east, west, north, or south.
 * 2) The direction of a vector is expressed as a counterclockwise angle of rotation of the vector about its "__ [|tail] __" from 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 and etc. This is the most commonly used convention. Assume it is this option if not given direction: Always assume from East if not given.

Vectors are drawn to show magnitude of the movement of an object. They must include a scale and either an angle according to a Cardinal direction or an angle according to the counter-clockwise measurement from east.


 * Topic Sentence:** Vectors convey both magnitude and direction, which can be described in a variety of ways, and can be shown via scaled vector diagrams.

b)Vector Addition You can add vectors together; The sum is the __resultant__. The //net force// experienced, or the resultant by an object is determined by computing the vector sum of all the individual forces acting upon that object. Observe the following summations of two force 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** This is useful for determining the result of adding ONLY two vectors __that make a right angle__ to each other. 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. However, this method does not determine the direction of the vector; this is done through trig.

**Using Trigonometry to Determine a Vector's Direction** The direction of a //resultant// vector can be determined by use of trigonometric functions. SOH CAH TOA helps one remember the meaning of sine, cosine, and tangent functions. These three trigonometric functions can be applied to the hiker problem above to determine direction. You start by selecting on of the two angles (NOT the right angle) of the triangle. Once the angle is selected, use any function to find 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. 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.

The Pythagorean Theorem and Trig Method __ONLY__ works for problems involving __RIGHT ANGLES__. Any others are solved by using the head to tail 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 found using an accurately drawn scaled vector diagram. Using this, the **head-to-tail method** is employed to determine the vector sum/resultant.

__Head-To-Tail Method:__
 * 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 a protractor, then putting it in the counterclockwise convention from due East.

An example is shown below, adding three vector quantities. 20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg. SCALE: 1 cm = 5 m Its magnitude and direction is labeled on the diagram. SCALE: 1 cm = 5 m

__**//The order in which three vectors are added does not affect the magnitude and direction.//**__
 * Topic Sentence:** Vectors can be added to find a resultant magnitude and direction; the Pythagorean Theorem and trigonometry allow us to find the resultants of vectors that are at right angles to each other, and the head-to-tail method allows us to find the resultants of vectors that are at variegated angles.

Reading Notes: Vectors Lesson 1 (c,d)
c)Resultants Any two (or more) vectors can be added as long as they are the same vector quantity. For example, a velocity vector and a force vector cannot be added together.
 * Topic Sentence:** Resultants are the sum of adding two or more vectors, and can take the place of all of them combined.

d) Vector Components Some vectors are in //two dimensions// - upward and rightward, northward and westward, eastward and southward, etc. All vectors have two parts. Each part 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 displacement. The component of a single vector describes the influence of that vector in a given direction.


 * Topic Sentence:** Any vector in two dimensions has two different components.

Reading Notes: Vectors Lesson 1 (e)
e)Vector Resolution Any vector directed in two dimensions can be thought of as having two components. In this unit, we learn two basic methods for determining the magnitudes of the components of a vector directed in two dimensions. This process 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 its components. The step-by-step procedure above is shown 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 (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 method of employing trigonometric functions to determine the components of a vector are as follows: The above method is illustrated below. 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. 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 two methods of vector resolution; the parallelogram or trig method.
 * 1) Construct a //rough// sketch of the vector in the indicated direction. Label the magnitude and angle 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 tail of each component begins at the tail of the vector and stretches along the axes to the nearest corner of the rectangle.
 * 4) Label the components with symbols to indicate the components. A northward force component might be labeled Fnorth, 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.
 * Topic Sentence:** A vector can be resoluted using either the parallelogram or trig method.

Reading Notes: Vectors 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, 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 with respect to the observer on land will not be the same as the speedometer reading of the vehicle 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.

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? Since side winds form a right angle, you can simply use pythagorean theorem for the magnitude and the trigonometric method for the direction.

**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. The resultant velocity of the motorboat can be determined in the same manner as was done for the plane. 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 (shown in the example):  The 2nd and 3rd questions can be answered by using the average speed equation: The distance of 80 m can be substituted into the numerator. But how about the denominator? 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. The same equation must be used to calculate the //downstream distance,// or part C of the question. 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.
 * < 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. <-- (This is found using the pythagorean theorem)
 * 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?
 * ave. speed = distance/time** ||

Topic Sentence: Riverboat and Relative Velocity problems can be solved easily by the use of the avg. speed equation.

h) Independence of Perpendicular Components of Motion A diagonal vector has two parts- an upward direction and a rightward direction. These two parts of the two-dimensional vector are **components,** which describe the affect of a single vector in a given direction. The vector sum of these two components is always equal to the force at the given angle.

Any vector 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. 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." 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.

If a plane was flying northward in a crosswind to the east, then the resulting velocity would have two components; one to the east and one to the north. However, if the wind velocity changed, the plane would still have the same rate going in the northern direction; the alteration in wind velocity does nothing to affect the plane's northerly direction. This is true for riverboat problems as well; the current will not affect the speed of the motor of the boat or the time it will take to reach the other riverbank, only the distance downstream the boat will land.

All vectors can be thought of as having perpendicular components. 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.


 * Topic Sentence:** Perpendicular components of vectors have no affect upon the motion of the other.

Reading Notes: Chapter 2 (a,b,c)
a)What is a Projectile? The main idea of this chapter is to preview what projectile motion is and the fundamentals of projectile motion. Projectile motion is both the vertical motion and horizontal motion.

1)What is a projectile? It is any object that once //projected// or dropped continues in motion by its own __ [|inertia] __ and is influenced only by the downward force of gravity.

2)Why is inertia important for projectiles? Inertia is the tendency for things to continue its state of motion or rest. If an object is moving, then it will keep moving in the same direction unless a force changes its direction or speed. For projectiles, the only force affecting this inertia should be gravity.

3)What is a common misconception of projectile motion? One common misconception is that when an object is thrown, it has a continuous force affecting it in that direction. This is not true; a continuous force is not necessary to keep the object in motion. Gravity is the only force affecting a projectile; there is no force upwards and no force to whichever direction the object is being projected. Also, horizontal forces are not necessary to keep the object moving horizontally.

4)What are the types of projectile motion? A projectile can either be dropped, can be thrown up vertically, or can be thrown at an angle. The latter two form a parabolic shape.

5)How is projectile motion different from freefall? They are very similar, except projectile motion concerns motion in 2 dimensions whereas freefall is only in one direction.

b)Characteristics of a Projectile's Trajectory The main idea of this chapter is to provide examples of various types of projectile motion, and to give a clear understanding of it. Main points:
 * A projectile is any object upon which the only force is gravity,
 * Projectiles travel with a parabolic trajectory due to the influence of gravity,
 * There are no horizontal forces acting upon projectiles and thus no horizontal acceleration,
 * The horizontal velocity of a projectile is constant (a never changing in value),
 * There is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down,
 * The vertical velocity of a projectile changes by 9.8 m/s each second,
 * The horizontal motion of a projectile is independent of its vertical motion.

1)What will happen if there was no gravity and a cannonball was shot horizontally? Why? The cannonball would keep going horizontally because no other force acts upon it.

2)What happens to the cannonball's trajectory if we add gravity again? Why? The cannonball would go at the same rate horizontally but would also accelerate downwards due to the force of gravity.

3)Does the horizontal motion alter the vertical motion, and vice versa? No. Perpendicular components of motion are independent of each other. The object would continue to move at the same rate in the horizontal direction even if a vertical direction was added to it, and vice versa.

4)What is the motion if the cannonball was launched upwards at an angle between 0º-90º without gravity? With gravity? The trajectory would be continuous in that angle direction if there was no gravity. If there was gravity, the cannonball would continue going at the same rate towards that angle direction, but would also have the downward force of gravity pulling it down and ultimately changing the trajectory into a parabolic shape.

5)In projectile motion, what is the only force affecting on the projectile and which components does it affect? Only gravity affects projectiles, and it only affects the vertical component. The horizontal component will remain the same; gravity has no affect on it.

c)Describing Projectiles With Numbers: (Horizontal and Vertical Velocity) The main idea of this reading is how to describe and find horizontal and vertical velocity and also the components of an object in projectile motion. It also describes the displacement of a projectile.

1)Does the negative acceleration of gravity affect the horizontal motion of a projectile? No, if an object goes 20 m/s horizontally then it will keep doing so. Only the vertical motion is affected.

2)If the object was launched upward and rightward, how would the horizontal and vertical velocities change over time? The horizontal velocity would remain the same, but the vertical velocity would change. The object would slow down as it reached its peak and would then speed up as gravity exerted its force upon it. For the initial velocity, it is necessary to compute the x and y components (for example, an object launched 75.7 m/s at a 15º angle would have an x-component of 73.1 m/s and a y-component of 19.6 m/s.

3)How does velocity and displacement on the way up compare with velocity and displacement on the way down? The trajectory of the projectile is symmetrical; the speed at one point on the way up will be the same speed at the same point on the way down. The velocity will have the same magnitude, but the object will be traveling in the opposite direction (+ on the way up, - on the way down). The displacement on the way up and down are also symmetrical.

4)What are the vertical and horizontal equations for a dropped projectile? What does this look like?
 * Vertical equation: y = 0.5 • g • t2**
 * Horizontal equation: x = vix • t**

5)How do you find the displacement of a projectile launched at an angle? What does this look like? The vertical displacement is the same equation as question 4, but with the initial velocity added on. The equation for horizontal displacement stays the same.
 * Vertical Displacement: y = viy • t + 0.5 • g • t2**
 * Horizontal displacement: x =** **vix • t**

Class Activity: Ball-in-Cup 9/24/11-9/25/11
Rob Kwark, Matt Ordover, Ryan Luo

media type="file" key="Horizontal Projectile Motion.mov" width="300" height="300"

Data: Average final distance: 3.16 m
 * ||  || Distance To Front Of Launcher || Final Distance ||
 * 1 || 3.032m || +8 cm || 3.112 m ||
 * 2 || 3.038m || +8 cm || 3.118 m ||
 * 3 || 3.102m || +8 cm || 3.182 m ||
 * 4 || 3.098m || +8 cm || 3.178 m ||
 * 5 || 3.096m || +8 cm || 3.176 m ||
 * 6 || 3.100m || +8 cm || 3.180 m ||

a) How fast does the launcher shoot the ball at "medium range"? It shoots, on average, about 7.07 m/s b) After changing the initial height of the launcher, calculate where to place the cup on the floor so that ball lands inside of the cup 3 times in a row. The cup should be 2m 69.4 cm away from the launch point. c) Calculate the %error of the theoretical position of the cup with the actual position of the cup. There was a 0% because the theoretical position of the cup, 2m 69.4 cm, was indeed the point where the ball landed inside the cup. However, the launcher was not very precise because it would occasionally shoot a bit too far or a bit too short. If we must calculate % error, it should be the % error of the launcher, not the cup, but we do not know the precise standard shooting speed of the device (which would be the theoretical) so we cannot determine this value. We got the ball inside the cup in about 60% of the time (around 6 out of 10 tries).

Gourd-O-Rama:
Picture: Materials: Cardboard box, tile, skateboard wheels and axles, paper, marker, tape, gourd.

Procedure: We nailed the axles through the bottom of the cardboard box and the tile, then put paper around the box and colored it.

Analysis (Initial velocity and Acceleration): Time: 3.93 seconds Distance: 9.2 m Initial Velocity: 4.682 m/s Acceleration: -1.19 m/s^2

Conclusion: Our project did pretty well, but we could have made it much better. When the contraption reached the bottom, the weight of the gourd inside tipped it over until we taped it securely. And when it finally started rolling, the bolts fell off, then a wheel, which made it gradually turn right and hit the wall, stopping it a lot earlier than it should have stopped. We could have fixed this by making sure the nuts, bolts, and wheels were more secure.

Orienteering Lab (10/14/11)
Lab Partners: Matt Ordover, Ryan Luo

Vectors: Actual Resultant: 26.40 m
 * Leg A: || North || 8.24 m ||
 * Leg B: || West || 9.26 m ||
 * Leg C: || North || 4.28 m ||
 * Leg D: || West || 10.17 m ||
 * Leg E: || North || 5.20 m ||

Graph Diagram: Measured Distance (on graph) - 26.67 m Measured Angle (on graph) - 66º W of N

Analytical Method: Calculated distance: 26.30 m Calculated Angle: 65.8º

Percent Error: ---Of Calculated to Actual (experimental): (Actual value - Calculated Value) / (Actual value) x 100 (26.40 - 26.30) / 26.40 x 100 0.379% error ---Of Measured to Actual (experimental): (|26.67 - 26.4|) / 26.67 x 100 1.01% error

The measured value had a higher percent error than the calculated value, probably due to the fact that I was a tiny bit off with the length of each leg when I was drawing; even half of a millimeter off could be significant because of the scaling.

Football Field Orienteering (10/21/11)
Walking (actual) displacement: 67.16 m.


 * Leg A: || North || 9.22 m ||
 * Leg B: || East || 24.52 m ||
 * Leg C: || North || 18.31 m ||
 * Leg D: || East || 24.23 m ||
 * Leg E: || North || 18.33 m ||

Graph Diagram:

Measured Distance and angle (on graph): 67.2 m, 43.2º

Analytical Method: Calculated Displacement: 66.9 m, 43.25º

Analysis: Percent Errors:

The graphical percent error was a bit lower than the % error of the analytical; however, both remained very close to 0% error. This may be due to the fact that while we were measuring the actual distances outside, we may have been off by a few centimeters.

Lab: Shoot Your Grade
Lab partners: Matt Ordover, Ryan Luo 11/4/11

**Purpose with Rationale**: Our purpose is to be able to calculate the trajectory of a ball using a launcher with a set velocity and a given angle, which was in our case 15º. This would be an off-the-cliff type of problem. In order to test our trajectory, we would set up 5 rings and a cup at specific distances away from the launches, finding the times and the y-values at which the rings needed to be placed. My hypothesis is that the ball will follow a curved trajectory, for which we will be able to find position at variable times; this will enable us to set up the rings and cup in a manner that will allow the ball successfully pass through all the rings and into the cup.

Materials Used: We used tape rolls as the rings, string, projectile launchers, little balls as projectiles, clamps, duct tape (to hold the string and tape), paper, carbon paper, measuring tape, binder clips, black stick (for loading the launcher) and a yard stick.

Method (Procedure): First, we set the launcher up on a ledge, securing it with two clamps. Then we shot the projectile at medium speed, positioning the carbon paper and paper near the point where the projectile would hit. After doing this 6 times, we measured the dots' distances away from the launcher and took the average. Then we calculated the initial velocity of the launcher using this average distance away from the launcher (or the x-value) so that we could use the value in following calculations. Afterwards, we measured the height of the cup + the projectile (so that the ball would be just a bit above the cup; we didn't want to make the ball hit the edges of the cup) using the tape measure and used that to calculate its distance away (x-value) from the launcher. Then we used the already-hanging tape rings from the previous class and measured their distances away from the launcher (the x-values); we used that to calculate the y-value of the rings at each distance. We set the rings up by putting two strands of string on each roll of tape and hanging them up from the ceiling tile edges. We made sure that all the rings were exactly at the theoretical/calculated height and distance away from the launcher. We shot the launcher 13 times, each with varying results; the amount of rings that the projectile went through varied on each test. After those test runs, we placed the cup in its position and called Mrs. Burns over, and after 4-5 tries made it through all 5 rings and the cup.

**Horizontal Range and Vertical Height of Launcher:** -To find the horizontal distance, we set up carbon paper and regular paper under it. Then we shot the projectile and measured the distance of the carbon points away from the desk with a tape measure. *We added 11.5 cm because the launcher nozzle was that far away from the edge of the table. We subtracted 10 cm because we began measuring from the 10 cm mark on the tape measure for more accuracy.
 * || Initial Distance || 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 ||

-We measured the height of the launcher's nozzle to be 1.172 m

**Initial Velocity Calculations:** Equation for x: x: d=vt+1/2at^2 4.838=Vcos(15º)t V=4.838/(cos(15º)t)
 * || x || y ||
 * V(initial): || Vcos(15º) || Vsin(15º) ||
 * a: || 0 || -9.8 ||
 * t: || t= .71 s || t= .71 s ||
 * d: || 4.838m || -1.172 m ||

Equation for y: y: d=vt+1/2at^2 -1.172=Vsin(15º)t - 4.9t^2

Plug x-equation into the y-equation to find t: -1.172=(4.838/(cos(15º)t) x (sin(15º)t - 4.9t^2 t=0.71 sec

Plug t back into the x-equation to find initial velocity: V=4.838/(cos(15º)t) V=7.054 m/s

cup height: .115 m Total change in distance: -1.172 m + .115 m = -1.057 m =6.814 || 7.054sin(15º) =1.826 || We are looking for the horizontal distance away, but in order to do so we need to find time first. d=Vt+1/2at^2 -1.057=1.826t - 4.9t^2 t=0.687 seconds
 * Cup Calculations:**
 * || x || y ||
 * V(initial) || 7.054cos(15º)
 * a: || 0 || -9.8 ||
 * t: || t= .687 || t= .687 ||
 * d: || d= 4.68 m || d=-1.057 m ||

Plug t into the equation for horizontal motion: d=Vt+1/2at^2 d=(0.687) x (6.814) d=4.68 m away

Ring 1 @ 1.10 m away from the launcher We need to find the vertical distance from the ground to know where place the ring, but first we need to find the time. d=Vt+1/2at^2 1.10=6.814t t=0.1614 sec
 * Ring Calculations (1 example calculation is provided):**
 * || x || y ||
 * V(initial) || 6.814 m/s || 1.826 m/s ||
 * a || 0 || -9.8 ||
 * t || t= .1614 || t= .1614 ||
 * d || 1.10m || d=. 1671m ||

Now plug that into the equation for y to find vertical distance: d=Vt+1/2at^2 d=(1.826)(.1614) - 4.9(.1614)^2 d=.1671 m <--- But this distance means .1671 m ABOVE the launcher level. So add .1671 m + 1.172 = 1.3391 m off the ground. Or, 133.91 cm off the ground.


 * Calculations for all rings and cup:**
 * || Time (in seconds): || Horizontal Distance Away (x value): || Height of ring (y value): ||
 * Ring 1 || .1614 || 1.10 m || 133.91 cm ||
 * Ring 2 || .2539 || 1.73 m || 131.97 cm ||
 * Ring 3 || .3317 || 2.26 m || 123.88 cm ||
 * Ring 4 || .3977 || 2.71 m || 112.32 cm ||
 * Ring 5 || .4726 || 3.22 m || 94.07 cm ||
 * Cup || .6870 || 4.68 m || 11.50 cm ||

Example: Ring 1 (y value) (Theoretical - Actual)/Theoretical x100 (133.91 - 133.91 / 133.91) x 100 = 0% 0% error.
 * Percent Error Calculations:**

Ring 1 (x value) (theoretical - actual)/theoretical) x 100 (1.1 - 1.1)/ 1.1 x 100 = 0% 0% error As outrageous as it seems, we didn't change the position of the rings. The little adjustments we made during class were us making sure the actual positions were exactly what the theoretical positions were. Once we got that perfect (and the rings all on the same path), we didn't deviate from the theoretical positions. The error was with the launcher; sometimes it shot the ball too fast, too slow, sometimes it was angled wrong vertically, sometimes it was pointed a bit to the right or left, all of which would ruin its chances of going through the rings and into the cup.
 * <  ||< Percent Error (x) ||< Percent Error (y) ||
 * < Ring 1 ||< 0% ||< 0% ||
 * < Ring 2 ||< 0% ||< 0% ||
 * < Ring 3 ||< 0% ||< 0% ||
 * < Ring 4 ||< 0% ||< 0% ||
 * < Ring 5 ||< 0% ||< 0% ||

Video/Procedure: media type="file" key="Shoot Your Grade.mov" width="300" height="300" Performance results:
 * || Number of times it occurred ||
 * Went through 2 rings || 2 ||
 * Went through 3 rings || 2 ||
 * Went through 4 rings || 3 ||
 * Went through 5 rings || 2 ||
 * Went through 5 rings + the cup || 1 ||

Conclusion: My hypothesis was correct. The ball launched at 15º at a speed of 7.054 m/s did follow a curved trajectory, reaching its maximum height either between the launcher and the 1st ring or between the 1st and 2nd ring. We can tell this because the height of the 2nd ring was slightly lower than the height of the 1st ring. By using various formulas, we were able to calculate the vertical distance that the rings needed to be at a certain horizontal distance and the horizontal distance the cup needed to be at a certain vertical distance. Visually, the ball did have a parabolic trajectory, at first slightly curving upwards and then curving downwards into the cup. We did not get much error in our calculation of where the ball should be at a certain point. Actually we had 0% error/deviation from our theoretical calculations for where the rings and the cup should be placed. We just had to make sure that the rings were all in a straight path for the ball to follow; our horizontal and vertical calculations were dead on. However, the problem lay with the launcher itself. The launcher was not perfectly precise in its launching; every time the velocity was a bit different. This would cause each and every run to be a bit different at random, making some failures and other successes. In addition, the launcher sometimes was not pointing perfectly at the path the rings made. This was an issue because this made the ball hit the side of the rings. Also, the angle the launcher was set at sometimes changed, changing the entire trajectory. We resolved the last two issues by making sure the clamps were extra tight and by placing duct tape on the back of the launcher to hold it in place (because the launch of the ball would make the launcher move a bit and/or jump up a bit, changing its angle and its position). Unfortunately, there is no way to make sure the launcher shot at exactly the same speed every time; that is why the ball had varying measures of success on each test run. A real life application of this whole project would be when a person is shot out of a cannon at a circus performance. The person would have to pass through a certain number of flaming rings and land in a pit of water; if they don't... well let's just say it won't be a happy ending for the person being shot. The cannon would have nearly the same sources of errors as our experiment.