Ch3_WangS

toc
 * Chapter 3: Vectors: Motion and Forces in Two Dimensions**

A) Vectors and Direction
[|Link] Quantities that describe the physical world fall into two categories- vector and scalar. A vector quantity is a quantity that is fully described by both magnitude and direction. A scalar quantity is a quantity that is fully described by its magnitude. Examples of vector quantities: displacement, velocity, acceleration, and force. These are respresented by scaled vector diagrams. They depict a vector by use of an arrow drawn to scale in a specific direction. Vector diagrams are often called free- body diagrams.
 * Topic Sentence:** Vectors are quantities with a magnitude and direction are shown on diagrams and have a head and tail and are drawn to scale.

A scaled vector diagram has:
 * 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).

**C****onventions for Describing Directions of Vectors** Vectors can be directed to each cardinal direction, such as north and northeast. Two methods for describing direction when it's not North, West, South, East are: > These show the second convention:
 * 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).
 * 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 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. 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.

B) Vector Addition
Two vectors can be added together to determine the resultant.The net force was the resultant of adding up all the force vectors. [|Link] Two methods for determining the magnitude and direction of the result of adding two or more vectors are:
 * Topic Sentence:** The three methods used to add vectors are: the Pythagorean theorem, with trigonometry, and with the head to tail method.
 * the Pythagorean theorem and trigonometric method
 * 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.

 **Using Trigonometry to Determine a Vector's Direction**   Once the measure of the angle is determined, the direction of the vector can be found. 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__.

**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 i s employed to determine the vector sum or resultant.

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. The process is repeated for all vectors that are being added. The resultant is then drawn from the tail of the first vector to the head of the last vector. 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.

C) Resultants
[|Link] The resultant is the vector sum of two or more vectors. **A + B + C = R**
 * Topic Sentence:** The resultant is the sum of separate vectors and is found by adding the individual vectors together using vector addition.

When displacement vectors are added, the result is a //resultant displacement//. If two or more velocity vectors are added, then the result is a //resultant velocity//. This applies to all different kinds of vectors.

D) Vector Components
[|Link]
 * Topic Sentence:** Any vector with two different directions would be equal to two different vector components in those two directions.

 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, or components. 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. The combined influence of the two components is equivalent to the influence of the single two-dimensional vector.

E) Vector Resolution
[|Link]
 * Topic Sentence:** The two different ways to find vector resolution are the parallelogram method and the trigonometric method.

That is, any vector directed in two dimensions can be thought of as having two components. Two basic methods for determining the magnitudes of the components of a vector directed in two dimensions are:
 * the parallelogram method
 * the trigonometric method

The process of determining the magnitude of a vector is known as vector resolution.

**Parallelogram Method of Vector Resolution** To use this method:  **Trigonometric** **Method of Vector Resolution** Here are the steps: 
 * 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 to form a rectangle.
 * 3) Draw the components, 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. Put arrows on these components to indicate tail/head.
 * 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.
 * 1) Construct a //rough// sketch 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 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.
 * 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.

**F) Component Method of Vector Addition**
[|Link]
 * Topic Sentence:** Since you can't use the Pythagorean theorem to solve all vector addition problems, you should use trigonometry to find the x and y components of vectors that are not at 90 degree angles.


 * Addition of Three or More Right Angle Vectors **

If you wanted to find the resultant for this equation:



This is a correct way of drawing a diagram.



This is also a correct way of drawing a diagram. 

Important point about adding vectors: the resultant is independent by the order in which they are added. Adding vectors **A + B + C** gives the same resultant as adding vectors **B + A + C** or even **C + B + A**. As long as all three vectors are included with their specified magnitude and direction, the resultant will be the same.


 * SOH CAH TOA and the Direction of Vectors**

The direction of a vector can be determined in the same way that it is always determined - by finding the angle of rotation counter-clockwise from due east. Since the resultant is the hypotenuse of a right triangle, this can be accomplished by first finding an angle that the resultant makes with one of the nearby axes of the triangle.


 * Addition of Non-Perpendicular Vectors**

The Pythagorean theorem is not applicable. Two non-perpendicular vectors will not form a right triangle. Yet it is possible to force two (or more) non-perpendicular vectors to be transformed into other vectors that do form a right triangle. The trick involves the concept of a vector component and the process of vector resolution.  A vector component describes the effect of a vector in a given direction. Any //angled// vector has two components; one is directed horizontally and the other is directed vertically.

Example:
 * A: 2.65 km, 140° CCW**
 * B: 4.77 km, 252° CCW**



**G) Relative Velocity and Riverboat Problems**
[|Link]
 * Topic Sentence:** Sometimes, the motion of an object is relative to where the observer is standing. Also, vector addition can be applied to riverboat problems, where the boat speed is affected by the current.

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 with respect to the observer on land will not be the same as the speedometer reading of the vehicle.

Consider a plane flying amidst a tailwind, a wind that approaches the plane from behind with an increase in its resulting velocity.

Since a headwind is a wind that approaches the plane from the front, such a wind would decrease the plane's resulting velocity. This is what would happen if there was a side wind:  You can use the Pythagorean theorem to add the vectors. You can then determine the angle of the vector using trigonometry.

 **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 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.So you can use the Pythagorean theorem to get the resultant velocity. Then, use trig to find the angle of the resultant.

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.
 * 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?

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. 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. 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.

H) Independence of Perpendicular Components of Motion
**Topic Sentence:** Two perpendicular components, which make the legs of a right triangle, make up the resultant vector. The individual components do not affect each other. [|Link] A force vector that is directed upward and rightward has two parts - an upward part and a rightward part. 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 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. 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 the boat is the combination 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. **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.

Orienteering Activity Part 1
Data for Orienteering: **Begin Position:** The second column on the left (when coming out of the science wing) under the overpass. **End Position:** The first column on the left (when coming out of the science wing) under the overpass. **Measured Displacement Resultant:** 760 cm

Scale Drawing of Orienteering:

Displacement result:



3.5 cm on scale drawing = 700 cm

Calculations for the Resultant: 613 cm N +900 cm S +360 cm S = 647 cm S 2690 cm W + 2655 cm E = 35 cm W

Actual (Experimental Value): 760 cm (7.6 m) Measured (Theoretical Value 1): 700 cm (7.0 m) Calculated (Theoretical Value 2): 647.95 cm (6.48 m)

Percent Error for Measured Resultant: Percent Error = [(theoretical-experimental)/theoretical]*100 Percent Error = [(700-760)/700]*100 Percent Error = 8.57%

Percent Error for Calculated Resultant: Percent Error = [(theoretical-experimental)/theoretical]*100 Percent Error = [(25.88-25.3)/25.88]*100 Percent Error = 17.23%

When I used the graphical method, as opposed to the analytical method, I got more accurate results. This is surprising because in class, we learned to use the analytic method more often because it is supposed to be more accurate. As always, there are multiple sources of error. When we were measuring the distances outside, there were no lines on the ground to guide us, so we probably didn't measure at precise 90 degree angles all the time. Also, the day we went out do this activity, it was raining and there was mud on the ground. As a result, we had to lift the tape measure off the ground to measure so of the distances. This could have caused us to get flawed data. It is obvious that we erred a lot in this activity, though, because of our high percent error. The good thing, however, is that in part two of this activity, we improved our error of margin considerably.

Orienteering Activity Part 2

 * Data for Orienteering: **



**Measured Displacement Resultant:** 55.85 m


 * Scale Drawing of Orienteering with Displacement: **



27.45 cm on scale drawing = 54.9 m
 * Resultant:**

9.50m E + 22.0m E + 16.27m E = 47.77m E 10.0m N + 17.22m N = 27.22m
 * Calculations for the Resultant: **



Actual (Experimental Value): 55.85 m Measured (Theoretical Value 1): 54.0 m Calculated (Theoretical Value 2): 54.98 m

Percent Error for Measured Resultant: Percent Error = [(theoretical-experimental)/theoretical]*100 Percent Error = [(54-55.85)/54]*100 Percent Error = 3.43%

Percent Error for Calculated Resultant: Percent Error = [(theoretical-experimental)/theoretical]*100 Percent Error = [(54.98-55.85)/54.98]*100 Percent Error = 1.58%

Our percent errors were 3.43% and 1.58%, both of which are less than 5%. Our results for this lab were pretty good, considering my percent errors for both parts was less than 5%. However, some errors were made, considering the percent error is not 0%. One source of error was the tape measure. It was not long enough to measure our whole resultant, so we had to measure it twice. This could have caused us to measure the resultant in an crooked line. To fix this, we need to get a longer tape measure. Also, we could have gotten erroneous results because we used the first 10 cm on the tape measure. Since a piece of black plastic covers the first 10 cm of the tape measure, it is impossible to know where the 0 cm mark actually is. To fix this, we could have just started measuring at the 10 cm mark, and then subtracting 10 cm from our result. But overall, I am happy with the results of this activity, as well as our considerable improvement from the last lab.

A) What is a Projectile?

 * Central Idea/Theme:** The main idea of the passage is how projectiles are similar to free fall objects. The two objects are similar because they are both only influences by gravity.. However, projectiles have a horizontal component as well as a vertical component.
 * 1) What is a projectile?
 * A projectile is an object upon which the only force acting is gravity.
 * A projectile is any object that once //projected// or dropped continues in motion by its own inertia and is influenced only by gravity.
 * 1) How is a projectile different from a free fall object?
 * A projectile is essentially an object in free fall that goes a vertical distance as well as a horizontal distance.
 * 1) What is inertia?
 * The resistance an object has to a change in a state of motion
 * 1) What is Newton's Law of Inertia?
 * An object in motion will continue to move in a straight line at constant speed if gravity was turned off.
 * 1) What are the different kinds of projectiles?
 * The first one is an object dropped from rest. The second one is an object thrown vertically upward, and the third is an object thrown upward at an angle to the horizontal. In all cases, the influence of air resistance is negligible.

B) Characteris**tics of a Projectile's Trajectory**

 * Central Idea/Theme:** Because of the influence of gravity, projectiles have a parabolic trajectory. The x and y components are independent of each other, however, and the x acceleration is 0 at all times.
 * 1) Does gravity affect the horizontal motion of a projectile?
 * No, gravity only affects the vertical motion of a projectile.
 * 1) Are the x and y components of a projectile's trajectory related?
 * No, they are independent of each other.
 * 1) What does the trajectory of a projectile look like?
 * It is parabolic.
 * 1) Is horizontal acceleration present?
 * No, the horizontal aspect of the projectile stays at constant speed.
 * 1) Is vertical acceleration present?
 * Yes, it is always 9.8 m/s/s downward.

C) Describing Projectiles with Numbers

 * Central Idea/Theme:** A projectile has a horizontal and vertical component. To find each, you should make a table including Vi, a, t, and d. You can find different components of projectiles by using a set of simple equations.
 * 1) What is the equation to find the vertical displacement of a projectile?
 * y = .5 * g * t^2
 * 1) What is the equation to find the vertical displacement of a projectile?
 * x = vix • t
 * 1) What is the equation for vertical displacement for an angled- launched projectile?
 * y = viy • t + 0.5 • g • t2
 * 1) When does a projectile speed up, and when does it slow down?
 * As it goes towards its peak, it slows down, and as it goes towards the ground, it speeds up.
 * 1) Is the vertical velocity ever zero?
 * Yes, at the peak of the path, the y velocity is zero.

Ball in Cup Lab

 * Part 1: Finding the initial velocity**

The ball went into the cup when the cup was 2.12 m away from the launcher.
 * Part 2: Using the initial velocity to get the ball into a cup off a different y height**

Percent Error = [(theoretical-experimental)/theoretical]*100 Percent Error = [(2.07- 2.12)/2.07]*100 Percent Error = 2.42%

I got a 2.42% percent error. Although I expected the ball to land in the cup when it was 2.07 m away, it actually landed in the cup when it was 2.12 m away. This margin of error was probably due to a few factors. First, the launcher is not always consistent, and does not launch the ball the same distance all the time. Also, the ball could have been launched differently each time, because the person launching the ball could have pulled the string harder sometimes, and softer other times. Also, the string could have been pulled to the side instead of straight up, altering the results. To eliminate these sources of error, we should do multiple trials to get the best result. Also, we should assign one person to pull the string of the launcher straight up each time. If only one person pulled the string, there's a better chance that the string would be pulled with the same strength each time.

Gourd-O-Rama Project

 * Partner: Hella Talas**
 * Picture of final project:**



d = 11.0 m t = 4.6 s vf=0 vi = ? a = ?
 * Data:**


 * Calculations:**

a= -1.04 m/s/s vi= 4.78 m/s
 * Results:**

The main problem with our cart was that the wheels would bend under the weight of the pumpkin. We should have used scooter wheels, or wheels from a toy car instead of making them from from candle tin lids because they probably would have been sturdier. Also, since the body of our cart is made of cardboard, it is not that sturdy. To remedy this, we should have have reinforced the sides of our cart with wood or plastic. Our pumpkin didn't really fit in our cart, so we should have made it bigger, too. However, we also could have used a smaller pumpkin. I am satisfied with our project, however, as it did travel a fair distance, and had excellent acceleration and velocity results.
 * How would I make it better?**

Shoot Your Grade Lab

 * Partners**: Michael Poleway, Kosuke Seki


 * Hypothesis:** The ball will go through all five hoops and go into the cup. This is what should happen if we do our calculations correctly.


 * Purpose with Rationale:** For this lab, we launched a ball with a launcher at a 25 degree angle. The launcher was set on the counter, so this was an off the cliff problem. We wanted to arrange five hoops and a cup in a way so the ball would travel through the five hoops and land in the cup. For this lab, since we are on Earth, we are assuming that the y acceleration is -9.8 m/s/s because of the influence of gravity.


 * Materials:** The materials that we used for this lab were a launcher, a ball, a cup, tissur to put in the cup to cushion the ball, rolls of masking tape, masking tape, carbon paper, paper, string, measuring tape, meter stick, and ruler.

The first thing we did was calculate the initial velocity. We did this by setting up the launcher at a 25 degree angle and launching the ball. We noted where the ball landed, and set up a piece of carbon paper with a piece of normal paper over it. We taped this down, and launched the ball 5 more times. We measured the distances from the launcher to the different marks on the carbon paper (that were made when the ball hit), and took the average distance, which happened to be 3.21 m. We used this value as our total x distance. We then calculated the initial velocity. This value was 4.81 m/s. Since our launcher was set at a 25 degree angle, we did vi*cos25 to find the x initial velocity, and vi*sin25 to find the y initial velocity. The results were: x initial velocity = 4.40 m/s and y initial velocity = 2.05 m/s. We used this information and the equation d = vit + 1/2at^2 to find the y distance. We subtracted the height of the cup from the total y distance, so the ball would go into the cup and not hit the bottom. Then, we set up the rings. The rings were already set up at distances of 50 cm, 100 cm, 151 cm, 208 cm, and 258 cm from the launcher. We used the value of vi and these distances to see where we should put the rings, vertically. After we did this, we tested the launcher to see if the ball would go through the rings and into the cup. It didn't on the first try, but we kept moving the rings so the ball would go through until the ball went through all five rings, and into the cup. As our ball went through all five rings and into the cup, I think we were ultimately successful (even if it took a while!). One thing we did that I found really helpful was putting someone's hand behind a ring if it was unclear whether it went through or not. If it hit the person's palm, then the ball wen through, and if it didn't hit the person's palm, then it was clear the ball didn't go through. This helped us know for certain if the ball went through the hoop or not, because it is sometimes unclear if the ball went through or not due to the ball's speediness.
 * Method:**


 * Here's a picture of our set-up with all the major parts labelled:**




 * Observations:**

__X Distance (or how far the ball goes horizontally)__**:**



Ideally, the launcher would be consistent, and the ball would go the same distance every time. But, the launcher is not consistent, and is sometimes off. We took the average of these five distances, and got 3.21 m. This is the x distance we used for our calculations.

__Distances to put hoops:__



We used our calculation of the initial velocity to find the vertical position of where to put the hoops. We could not control the horizontal distance, as the hoops had already been set up before we started the lab.

__Calculation chart for total distance and total time:__



//I know that the initial velocity for x is vi*cos25 and the initial velocity for y is vi*sin25. This is because our launcher is set at a 25 degree angle. I also know that x acceleration is 0 m/s/s because in projectiles, there is no horizontal acceleration. I also know that the y acceleration is -9.8 m/s/s because of the influence of gravity. My group measured the horizontal distance of the projectile by using carbon paper to mark where the ball landed and measuring the distance with a tape measure. We got the y distance by measuring the height of the counter that the launcher is sitting on. Then, we measured the distance from the place the ball launches from to the counter. We added these two distances together to get the y distance. We then used this information to calculate the time and initial velocity. That is how I got all the information in this chart.//

//**Observations and Data on Performance:**//

//__Video of best performance:__//

media type="file" key="Shoot your grade lab video.m4v" width="585" height="585"

//__Chart of results:__//



//**Calculations:**//


 * Error Analysis:**

__Percent Error:__



__Sample calculation:__ Percent Error = {(theoretical-experimental)/theoretical} * 100 Percent Error = {(16.9-21.5)/16.9} * 100 Percent Error = 27.2%

My hypothesis was that the ball would go through all five rings and into the cup. This was correct, as our projectile was able to go through the five rings and into the cup. However, we did not get this results solely because of our calculations. As you can see from our percent errors, we had to adjust the the rings to accommodate the ball. Like all projectiles, our ball had a parabolic trajectory. This is why we set up the rings in a parabolic shape. As always, we had a few experimental errors. The ring that was off the most was the first one, with a percent error of 27.2%. The cup, however, had a percent error of 0% because our theoretical distance and our experimental distance were the same. The rest of the rings had percent errors somewhere between 27.2% and 0%. This is pretty good. Our launchers were not consistent, and would sometimes shoot the ball too far left or too far right. This is a possible source of error. To fix this, we should eliminated trials where the ball did not go through the exact center of the ring. Also, every time we pushed the ball into the launcher, the launcher would move, causing the angle to change. We had to reset the angle every time we wanted to launch the ball. We could have set the angle wrong, which would contribute to a possible source of error. The launcher would move because of a broken screw. To fix this source of error, we simply needed to fix the broken screw, which would allow us to perform this experiment without the launcher moving. Also, we performed this lab under the assumption that the ball is a projectile. Since there is air resistance, gravity is not the only force acting on the ball, and it is therefor not a projectile. To mitigate the influence of air resistance, we should perform this experiment in a vacuum. This lab was important because there are many real life applications. Divers who jump off of diving boards must know the initial velocity and angle to push off at**.** They must also know how far, horizontally, to jump so they do not go too far and end of diving onto land!
 * Conclusion:**