Ch3_WeinfeldM

=Chapter 3 = toc

October 12: Summarize Vectors, Lesson 1 a + b, Method 1
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. Vector quantities are often represented by scaled [|vector diagrams]. Vector diagrams depict a vector by use of an arrow drawn to scale in a specific direction. Vector diagrams were introduced and used in earlier units to depict the forces acting upon an object. Such diagrams are commonly called as [|free-body diagrams]. Vectors can be directed due East, due West, due South, and due North. But some vectors are directed //northeast// (at a 45 degree angle); and some vectors are even directed //northeast//, yet more north than east. 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 direction of a vector is often expressed as an angle of rotation of the vector about its " tail " from east, west, north, or south.
 * The direction of a vector is often expressed as a counterclockwise angle of rotation of the vector about its " tail " from due East.

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. Two vectors can be added together to determine the result (or resultant). The [|net force] was the result (or [|resultant] ) of adding up all the 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 is a useful method for determining the result of adding two (and only two) vectors that make a right angle to each other. The method is not applicable for adding more than two vectors or for adding vectors that are not at 90-degrees to each other.The direction of a //resultant// vector can often be determined by use of trigonometric functions. Most students recall the meaning of the useful mnemonic SOH CAH TOA from their course in trigonometry. These three functions relate an acute angle in a right triangle to the ratio of the lengths of two of the sides of the right triangle. The ** sine function ** relates the measure of an acute angle to the ratio of the length of the side opposite the angle to the length of the hypotenuse. The ** cosine function ** relates the measure of an acute angle to the ratio of the length of the side adjacent the angle to the length of the hypotenuse. The ** tangent function ** relates the measure of an angle to the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. The 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. The head-to-tail method involves [|drawing a vector to scale] on a sheet of paper beginning at a designated starting p osition. Where the head of this first vector ends, the tail of the second vector begins (thus, //head-to-tail// method). The process is repeated for all vectors that are being added. Once all the vectors have been added head-to-tail, the resultant is then drawn from the tail of the first vector to the head of the last vector; i.e., from start to finish. Once the resultant is drawn, its length can be measured and converted to //real// units using the given scale. The [|direction] of the resultant can be determined by using a protractor and measuring its counterclockwise angle of rotation from due East.

October 13: Summarize Vectors, Lesson 1 c + d, Method 1
The ** resultant ** is the vector sum of two or more vectors. If displacement vectors A, B, and C are added together, the result will be vector R. 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." Vectors are sometimes 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. The single two-dimensional vector could be replaced by the two components. 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.

October 17: Summarize Vectors, Lesson 1 e, Method 1
Any vector directed at an angle to the horizontal (or the vertical) can be thought of as having two parts (or 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 and the trigonometric method. The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector. Briefly put, the method involves drawing the vector to scale in the indicated direction, sketching a parallelogram around the vector such that the vector is the diagonal of the parallelogram, and determining the magnitude of the components (the sides of the parallelogram) using the scale. If one desires to determine the components as directed along the traditional x- and y-coordinate axes, then the parallelogram is a rectangle with sides that stretch vertically and horizontally. 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. In conclusion, a vector directed in two dimensions has two components - that is, an influence in two separate directions. The amount of influence in a given direction can be determined using methods of vector resolution. Two methods of vector resolution have been described here - a graphical method (parallelogram method) and a trigonometric method.

<span style="color: #000080; font-family: 'Times New Roman',Times,serif; font-size: 1.3em;">October 18: Summarize Vectors, Lesson 1 g+h, Method 1
<span style="font-family: 'Times New Roman',Times,serif;">On occasion objects move within a medium that is moving with respect to an observer. In such instances, 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. Motion is relative to the observer. The observer on land, often named (or misnamed) the "stationary observer" would measure the speed to be different than that of the person on the boat. The observed speed of the boat must always be described relative to who the observer is. To illustrate this principle, consider a plane flying amidst a ** tailwind **. A tailwind is merely a wind that approaches the plane from behind, thus increasing its resulting velocity. The resultant velocity of the plane (that is, the result of the wind velocity contributing to the velocity due to the plane's motor) is the vector sum of the velocity of the plane and the velocity of the wind. This resultant velocity is quite easily determined if the wind approaches the plane directly from behind. If the plane encounters a headwind, the resulting velocity will be less than 100 km/hr. Since a headwind is a wind that approaches the plane from the front, such a wind would decrease the plane's resulting velocity. Now consider a plane traveling with a velocity of 100 km/hr, South that encounters a ** side wind ** of 25 km/hr, West. The resulting velocity of the plane is the vector sum of the two individual velocities. To determine the resultant velocity, the plane velocity (relative to the air) must be added to the wind velocity. This is the same procedure that was used above for the headwind and the tailwind situations; only now, the resultant is not as easily computed. Since the two vectors to be added - the southward plane velocity and the westward wind velocity - are at right angles to each other, the [|Pythagorean theorem] can be used. In this situation of a side wind, the southward vector can be added to the westward vector using the [|usual methods of vector addition]. The magnitude of the resultant velocity is determined using Pythagorean theorem. The direction of the resulting velocity can be determined using a [|trigonometric function]. Since the plane velocity and the wind velocity form a right triangle when added together in head-to-tail fashion, the angle between the resultant vector and the southward vector can be determined using the sine, cosine, or tangent functions. The tangent function can be used. Like any vector, the resultant's [|direction] is measured as a counterclockwise angle of rotation from due East. 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, 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: <span style="font-family: 'Times New Roman',Times,serif;">The first of these three questions: 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). <span style="font-family: 'Times New Roman',Times,serif;">** ave. speed = distance/time. **A force vector that is directed upward and rightward has two parts - an upward part and a rightward part. That is to say, if you pull upon an object in an upward and rightward direction, then you are exerting an influence upon the object in two separate directions - an upward direction and a rightward direction. These two parts of the two-dimensional vector are referred to as [|components]. A ** component ** describes the affect of a single vector in a given direction. Any force vector that is exerted at an angle to the horizontal can be considered as having two parts or components. The vector sum of these two components is always equal to the force at the given angle. 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. All vectors can be thought of as having perpendicular components. In fact, any motion that is at an angle to the horizontal or the vertical can be thought of as having two perpendicular motions occurring simultaneously. These perpendicular components of motion occur independently of each other. Any component of motion occurring strictly in the horizontal direction will have no affect upon the motion in the vertical direction. Any alteration in one set of these components will have no affect on the other set.
 * 1) <span style="font-family: 'Times New Roman',Times,serif;">What is the resultant velocity (both magnitude and direction) of the boat?
 * 2) <span style="font-family: 'Times New Roman',Times,serif;">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) <span style="font-family: 'Times New Roman',Times,serif;">What distance downstream does the boat reach the opposite shore?

<span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Vector Activity
<span style="font-family: 'Times New Roman',Times,serif;"> <span style="font-family: 'Times New Roman',Times,serif;">These were the legs given to us from the other group. We knew that we started at the railing near the stairs by the windows, but we had to find out where we would end.

<span style="font-family: 'Times New Roman',Times,serif;">After we physically walked the route and found our end spot, we measured the displacement and found that it was 22.99 meters. We then did an analysis:

<span style="font-family: 'Times New Roman',Times,serif;">Graphical Analysis: <span style="font-family: 'Times New Roman',Times,serif;"> <span style="font-family: 'Times New Roman',Times,serif;">The graphical analysis results: Resultant = 23.04 meters at 2 degrees.

<span style="font-family: 'Times New Roman',Times,serif;">Analytical Analysis: <span style="font-family: 'Times New Roman',Times,serif;"> <span style="font-family: 'Times New Roman',Times,serif;">The analytical analysis results: Resultant: 23.16 meters at 1.68 degrees.

<span style="font-family: 'Times New Roman',Times,serif;">Percent Error Calculations (measured compared to graphical and analytical)

<span style="font-family: 'Times New Roman',Times,serif;">

Conclusion: The purpose of this activity was to apply the graphical and analytical analyses to a real situation and determine if they really work. Our small percent error indicates for both the graphical and analytically methods indicate that both methods are accurate. The analytical analysis had a larger percent error than did the graphical analysis (0.73% v 0.22%). Usually, the graphical has more room for error because people can read things differently and even a small difference in a measurement can make a big difference. It is therefore strange that we got a larger percent error for the analytical analysis. However, this is possible because the theoretical may not have been perfect because we got that value using measuring tape. Perhaps we didn't pull the tape tight enough or we were slightly off in reading the values. To fix this problem, we could have been more precise when making the theoretical. Either way, the percent errors for both were very small so we know that both graphical and analytic analyses work and can be applied to real life situations.

<span style="color: #000080; font-family: 'Times New Roman',Times,serif; font-size: 1.3em;">October 19: Summarize Vectors, Lesson 2 a + b, Method 3
<span style="font-family: 'Times New Roman',Times,serif;">__ **A** __ <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Questions: <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Central Idea: <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Answers to Questions: <span style="font-family: 'Times New Roman',Times,serif;">__ **B** __ <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Questions: <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Central Ideas: <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Answers to Questions:
 * 1) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> What is a projectile? Give a few examples of projectiles.
 * 2) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> What is difference between the free body diagrams of projectiles moving downwards, upwards, upwards and rightwards, or downwards and leftwards.
 * 3) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> What does Newton's law address that can explain why objects can move upward and/or rightward even though the only force acting on them is gravity?
 * 4) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> What is inertia and why is it relevant in this section?
 * 5) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> What would happen to the motion of an object if there was no gravity?
 * <span style="font-family: 'Times New Roman',Times,serif;"><span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> The central idea of this passage is that a projectile is an object that continues its motion by inertia and is only influenced by the downward force of gravity. A major theme in the reading is also that a force does __ not __ affect motion, only acceleration.
 * 1) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> A projectile 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. Projectile are the most common example of an object that moves in two dimensions. An object dropped from rest is a projectile, an object thrown vertically upward is a projectile, and an object thrown upward at an angle to the horizontal is also a projectile.
 * 2) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> The free body diagrams all look the same regardless of the direction. It always looks like this: [[image:u3l2a3.gif]].
 * 3) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> Newton's laws suggest that forces are required to cause an acceleration (not a motion). When an object is moving upward, there is a downward force and therefore a downward acceleration so the object is moving upward and slowing down.
 * 4) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> Inertia is the resistance an object has to a change in its state of motion. Projectiles continue in motion by means of their own inertia. They are only influenced by gravity.
 * 5) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> The object would continue in the same motion and the same speed in the same direction. This is because there would be no force acting against the motion therefore, no acceleration, and no change in velocity.
 * 1) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> What are the two components of a projectile's motion? Are they independent or dependent of one another?
 * 2) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> If projectiles only have a vertical force acting upon them, how is there horizontal acceleration?
 * 3) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> What information must you look for for horizontally launched projectiles?
 * 4) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> How does a non-horizontally launched projectile travel when gravity is present?
 * 5) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> What is the difference between how a projectile would travel with and without gravity?
 * <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> This section discusses the horizontal and vertical motions of projectiles. Projectiles travel with a parabolic trajectory because gravity accelerates them downward from their otherwise straight-line. The force of gravity does not affect the horizontal velocity, only the vertical.
 * 1) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> Projectiles have both horizontal and vertical motion. They are perpendicular components and therefore, independent of each other so they can be discussed separately.
 * 2) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> The force of gravity acts downward and this vertical force acts perpendicular to the horizontal motion. It will not affect the horizontal motion because the perpendicular components are independent. Thus, the projectile travels with a constant horizontal velocity and a downward vertical acceleration.
 * 3) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> Are there any forces? If so, what direction are they? Is there acceleration? If so, what direction? Is there velocity? If so, is it constant or changing?
 * 4) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;"> They travel with a parabolic trajectory.
 * 5) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Without gravity, it would travel in a straight-line. The gravity acts as a downward acceleration so it becomes parabolic. The downward force and acceleration result in a downward displacement from the position that the object would be if there were no gravity.

<span style="color: #000080; font-family: 'Times New Roman',Times,serif; font-size: 1.3em;">October 20: Summarize Vectors, Lesson 2 c, Method 3
<span style="font-family: 'Times New Roman',Times,serif;">__ **A** __ <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Questions: <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Central Idea: <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Answers to Questions:
 * 1) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">What happens to the horizontal component during the course of trajectory and the vertical velocity changes 9.8 m/s every second?
 * 2) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">What happens if the projectile is launched upward at an angle to the horizontal?
 * 3) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">What equation can be used to find the vertical displacement of a horizontally launched projectile?
 * 4) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">What is the equation for the horizontal displacement of the projectile?
 * 5) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">What happens when the projectile falls below the gravity-free path by a vertical distance of 0.5 X g X t^2.
 * Again, the important concept depicted in the above diagram is that the horizontal velocity remains constant during the course of the trajectory and the vertical velocity changes by 9.8 m/s every second
 * 1) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">The horizontal component remains constant.
 * 2) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">There are x and y components because there is rightward and upward motion. Horizontal component remains constant.
 * 3) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">y = 0.5 X g X t^2
 * 4) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">x = Vix X t
 * 5) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">The gravity-free path is no longer a horizontal since the projectile is not launched horizontally.

<span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Horizontally Launched Projectiles Activity
<span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Maddy, Dani, Jenna, and John

<span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Objectives:
 * <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">measure the initial velocity of a ball
 * <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">apply concepts from two-dimensional kinematics to predict the impact point of a ball in projectile motion
 * <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">take into account trial-to-trial variations in the velocity measurement when calculating the impact point.

<span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Pre Lab Questions <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Part A: Find the initial horizontal velocity of the ball.
 * 1) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">If you were to drop a ball, releasing it from rest, what information would be needed to predict how much time it would take for the ball to hit the floor? What assumption must you make?
 * <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">You would need to know the distance from the dropping point to the floor. You can assume that the acceleration is -9.81 m/s/s, assuming that the only force acting on the ball is gravity. You can also assume that the initial velocity is 0 because you are at rest. You are also assuming that you only are looking for the vertical because it is a dropped free fall problem.
 * 1) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">If the ball in Question 1 is traveling at a known horizontal velocity when it starts to fall, explain how you would calculate how far it will travel before it hits the ground.
 * <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Because the horizontal velocity remains constant (because there is no horizontal acceleration) you can use d=v/t. You can use the time discovered in the question above and the given velocity and then have enough information to solve the problem.
 * 1) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">A single Photogate can be used to accurately measure the time interval for an object to break the beam of the Photogate. If you wanted to know the velocity of the object, what additional information would you need?
 * <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">If you are using the photogate to measure the horizontal velocity, then you need to know the horizontal distance between the starting point and the Photogate and then you can calculate the velocity.
 * 1) <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Write your procedure and get approval from Ms. Burns before you proceed any further!
 * <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">What data will you need to collect? Remember that you must run multiple trials. Keep in mind your end goal!
 * <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">We already know a lot of information: horizontal velocity = 0 m/s, horizontal acceleration = 0, vertical acceleration = -9.8 m/s/s, and the angle we are shooting from is 0 degrees. We also need to collect the distance that the the shooting machine is from the ground.
 * <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">How will you analyze your results in terms of precision and/or in terms of accuracy?
 * <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">We will perform multiple trials. If you do multiple trials and they all end up in the same spot then you know that your results are better than had you just done one trial. We can also analyze our results by using previously found information to find new data. For example, when we want to calculate where to put the cup, we can use what we found prior to the test to analytical get a distance. We can then test this and see if it works. A percent error calculation will then let us know how precise our values were.

<span style="color: #000080; font-family: 'Times New Roman',Times,serif;">

<span style="color: #000080; font-family: 'Times New Roman',Times,serif;">Part B: Change the initial height, calculate where to place the cup on the floor so that the ball lands inside of it three times in a row. Calculate percent error. <span style="color: #000080; font-family: 'Times New Roman',Times,serif;">

The work above did not account for the height of the cup. When we tried getting the ball in the cup, using the distance calculated before, it did not work. The ball hit the side of the cup instead of landing inside of it because the height of the cup had been ignored in our calculations. Therefore the ball was landing right before it got to the top of the cup. We then found that the cup was .095 m so we used this new information to give us a more accurate distance to place the cup. Work is below.

<span style="color: #000080; font-family: 'Times New Roman',Times,serif;">

Percent Error Calculation: Video: media type="file" key="working ball in a cup.m4v" width="300" height="300"

Conclusion: The purpose of this lab was to apply what we were learning to real life. We were trying to see if our calculations could be applied to set up a real experiment, and work accurately. We found that you can use these equations and because our percent error was only 0.481%, we know that they are very accurate. Of course, it was not perfect and there are several sources of error. The most likely problem was probably how we rounded and the fact that our measuring was not so precise. To fix this, we would need more precise measuring tools. There was also some error in lining the cup up perfectly lateral to the launcher. This is a difficult problem to fix, but perhaps we could have measured how far the launcher was from the wall and then measured how far our cup was from the wall to make sure they were completely lined up.

Shoot Your Grade Lab
Our given angle: 30 degrees Step by Step Procedure: 1. Calculate initial velocity from the ball in cup lab. 2. Use carbon paper and tape measure to see how far the ball goes. Divide length by 6 for the 5 rings and cup and then this is the horizontal distance at each of the 6 checkmarks. 2. Use given angle and initial velocity to perform off the cliff equations for the five different rings and the cup. 3. Set up rings, one at a time, and launch ball to see if it works. 4. Make necessary adjustments to each ring 5. Launch the ball through as many rings as possible and hope to get it in cup.
 * Objective**: The purpose of this lab was to set up five rings and a cup on the floor so that a ball shot out of the launcher at a given angle (30 degrees) and calculated initial velocity will pass through five rings and then land inside of a cup placed on the floor. We were supposed to use our calculations to place the rings and cup in the correct locations and the goal was to test how accurate projectile calculations are and if they can apply to real life situations.
 * Hypothesis**: We will be able to shoot the ball through all hoops and have it land in the cup based off of our calculations. The set up of the rings will start off with increasing in height and then getting shorter and shorter as the ball is a projectile and will have parabolic trajectory.
 * Methods and Materials:** We calculated the height using the off the cliff method. We then used our calculations to set up the rings. We measured the heights we got from the launcher, both horizontally and vertically, to line them up. We did set up one ring at a time and then would test it. They required a little bit of changing, but they were all in the right general area. Our materials were rings of masking tape, string, measuring tape, and carbon paper.

Pictures of Procedure: set launcher to 30 degrees measuring distance from ring to launcher using weight hanging rings testing hanging rings

media type="file" key="6 test run for procedure.mov"Procedure Video

Calculation for Initial Velocity: We used the five trials using carbon paper to find the average range of the ball.We then measured the horizontal height of the launcher and used these values to calculate the hang time and both components of the initial velocities. We found that the hang time was 0.795 seconds and that the x component was 4.2m/s and the y component was 2.43 m/s.

To calculate the horizontal distance we divided the range (3.342 m) by 6 (for the 5 rings and cup). Therefore, each ring was place 0.577m away from the previous.

Calculations for Vertical Distance (from launcher)



^ values above are in relation to the initial height, where the ball was launched from

media type="file" key="Movie+on+2011-11-02+at+09.40+
 * Procedure Video:**

The vertical values are in relation to the launcher to the hoops. The horizontal values are a straight path from to the hoop.
 * Measurements From When Rings Actually Worked**

media type="file" key="Shoot for Your Grade - 4 Hoops - Medium.m4v" width="300" height="300"
 * Video of Ball Getting Through 4 Rings!**

Our hypothesis was not completely correct. We had hypothesized that based off of our calculations, we would be able to set up the scene so that the ball would launch through all five rings and land through the cup. As is evident from our percent error calculations, we were unable to set it up perfectly based off of our calculations. A look at my percent error table illustrates why this part of our hypothesis was wrong. For example, we had predicted that each ring would be exactly 0.557 meters away from the one before it on the horizontal distance, but the rings ended up shifting forward or backwards. The vertical distances were different for all of the rings. For example, we had predicted that the first ring be placed 0.233 meters higher than the launcher but it needed to be 0.235 meters higher. These numbers are very close and all of the percent error calculations remained under 5% so our calculations were definitely very helpful. The other part of my hypothesis was about the placement of the rings in comparison to the others. I predicted that the rings would start off getting higher and then get lower, because projectiles travel with parabolic trajectory. This part of my hypothesis was proven true because the vertical distances increase from the launcher at first, and then get lower, and the negative values indicate that the rings got lower than the launcher.
 * Percent Error Calculations: (actual = experimental)**
 * Conclusion:**

The percent error occurred in both the horizontal and vertical distances. As stated before, the percent errors were all below 5%. Some were 0% and some were as close to 0% as 0.36%. The first ring had a horizontal percent error of 0. This is because we placed the first ring exactly where we had calculated it had gone. When it did not work out perfectly, we altered the vertical rather than the horizontal. Therefore, the vertical percent error was 0.86%. As we moved further along, the percent errors began to increase. For example, ring 3 had a horizontal percent error of 1.19% and a vertical percent error of 2.09%. We had a really hard time with the third ring because we had to set it up near the light bulb. Therefore we used the hooks instead of placing it through the ceiling. We tied the string around the hook, but it kept getting looser and would move slightly each time. Therefore, it was a challenge to get the perfect measure which explains why the percent error here is bigger. The percent error for the cup was hard to calculate because we did not get the ball to land in the cup. We were unable to find out where the cup should be placed horizontally because it wasn't landing where we had calculated it should. We were able to figure out the vertical distance of the cup by subtracting the height of the cup, 0.095 m from the distance of the launcher, 1.17 m. I could therefore calculate the percent error between this and what I got using the equation in my calculations and found a percent error of 0.47%. Overall, the percent errors came from inaccuracies in measuring as well as inconsistencies with the launcher. Several times, we would look at the degree on the launcher and realize that it moved. Even the slightest movement makes a huge distance.

To change the lab or make it more accurate, I would first make sure that the angle on the launcher was always exactly 30. Although we can't control all of the inaccuracies of the launcher, we can control what angle it is set to. I would also make sure that all of the strings are set up tightly so that they can not move at all. When they loosened even a little bit, the ball would hit the side of the ring, throwing everything else off. There is a lot of real life applications to this lab. First off, we learned about patience. This lab was very stressful and it was important to stay calm and focused. It also showed us the importance of consistency and attention to detail. In terms of physics, it showed us that the calculations we do in class can apply to real life situations, even if they do require a bit of an adjustment.

**Gourd O Rama**

 * Front View:**
 * Top View:**
 * Calculations:**


 * Results:** Our **velocity was 3.33 m/s** and our **acceleration was -0.888 m/s/s**! The cart and pumpkin **weighed a total of 3.2 kg** and **travelled 6.25 meters.**


 * Conclusion:** In order to improve our cart, the first thing I think we should work on is making it lighter. We used two layers of wood to ensure that the pumpkin would be stable, but I think one would have been enough. We could have also moved the wheels closer together or used smaller wheels. Another change that could have helped would be to change the shape of the cart. Had we made the front pointier and then gradually got wider, the velocity would have been faster. Our acceleration rate was really low so that wouldn't need as much work. We were happy with our overall distance, but obviously would like it to go even further. It kept turning towards the right so perhaps our wheels weren't lined up perfectly. Using axels would easily solve this problem. In conclusion, to improve our project, we would make a lighter, smaller project, with a pointed front. We would use axels to ensure the wheels were perfectly lined up and hope to improve our results!!