Ch3_SellmanJ

=**Method 1: Lesson 1 A and B**= toc

 __Lesson A__ 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 that have been [|previously discussed] include [|displacement], [|velocity] , [|acceleration] , and [|force]. 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. 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. Directions are described by the use of some convention. The most common convention is that the direction of a vector is the counterclockwise angle of rotation which that vector makes with respect to due East. __Lesson B__ Two vectors can be added together to determine the result (or resultant). For example, a vector directed up and to the right will be added to a vector directed up and to the left. There are a variety of methods for determining the magnitude and direction of the result of adding two or more vectors. The **Pythagorean theorem** is a useful method for determining the result of adding two (and only two) vectors __that make a right angle__ to each other. The method is not applicable for adding more than two vectors or for adding vectors that are __not__ at 90-degrees to each other. The Pythagorean theorem is a mathematical equation that relates the length of the sides of a right triangle to the length of the hypotenuse of a right triangle.The direction of a //resultant// vector can often be determined by use of trigonometric functions. 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 position. Where the head of this first vector ends, the tail of the second vector begins (thus, //head-to-tail// method). The process is repeated for all vectors that are being added. Once all the vectors have been added head-to-tail, the resultant is then drawn from the tail of the first vector to the head of the last vector; i.e., from start to finish. Once the resultant is drawn, its length can be measured and converted to //real// units using the given scale. The [|direction] of the resultant can be determined by using a protractor and measuring its counterclockwise angle of rotation from due East. A step-by-step method for applying the head-to-tail method to determine the sum of two or more vectors is given below.  The problem involves the addition of three vectors: ** 20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg. ** ** SCALE: 1 cm = 5 m ** The head-to-tail method is employed as described above and the resultant is determined (drawn in red). Its magnitude and direction is labeled on the diagram. SCALE: 1 cm = 5 m Interestingly enough, the order in which three vectors are added has no affect upon either the magnitude or the direction of the resultant. The resultant will still have the same magnitude and direction. For example, consider the addition of the same three vectors in a different order. ** 15 m, 210 deg. + 25 m, 300 deg. + 20 m, 45 deg. ** ** SCALE: 1 cm = 5 m **When added together in this different order, these same three vectors still produce a resultant with the same magnitude and direction as before (20. m, 312 degrees). The order in which vectors are added using the head-to-tail method is insignificant. SCALE: 1 cm = 5 m =**Method 1: Lesson C and D** = <span style="display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">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//. 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//. The resultant vector (whether a displacement vector, force vector, velocity vector, etc.) is the result of adding the individual vectors. <span style="font-family: Arial,Helvetica,sans-serif;"> A [|vector] is a quantity that has both magnitude and direction. Any vector directed in two dimensions can be thought of as having an influence in two different directions. That is, it can be thought of as having two parts. Each part of a two-dimensional vector is known as a ** component **. The components of a vector depict the influence of that vector in a given direction. The combined influence of the two components is equivalent to the influence of the single two-dimensional vector. The single two-dimensional vector could be replaced by the two components. Consider an airplane that is flying from Chicago's O'Hare International Airport to a destination in Canada. Suppose that the plane is flying in such a manner that its resulting displacement vector is northwest. If this is the case, then the displacement of the plane has two components - a component in the northward direction and a component in the westward direction. This is to say that the plane would have the same displacement if it were to take the trip into Canada in two segments - one directed due North and the other directed due West. If the single displacement vector were replaced by these two individual displacement vectors, then the passengers in the plane would end up in the same final position. The combined influence of the two components is equivalent to the influence of the single two-dimensional displacement. =<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Method 1: Lesson E = <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Any vector directed at an angle to the horizontal (or the vertical) can be thought of as having two parts (or components). That is, any vector directed in two dimensions can be thought of as having two components. For example, if a chain pulls upward at an angle on the collar of a dog, then there is a tension force directed in two dimensions. This tension force has two components: an upward component and a rightward component. The process of determining the magnitude of a vector is known as ** vector resolution **. The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector. This method involves drawing the vector to scale in the indicated direction, sketching a parallelogram around the vector such that the vector is the diagonal of the parallelogram, and determining the magnitude of the components (the sides of the parallelogram) using the scale. If one desires to determine the components as directed along the traditional x- and y-coordinate axes, then the parallelogram is a rectangle with sides that stretch vertically and horizontally. A step-by-step procedure for using the parallelogram method of vector resolution is: <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector. The method of employing trigonometric functions to determine the components of a vector are as follows: =Method 1: Lesson 1 G and H= <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">__Lesson G__ <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">A motorboat in a river is moving amidst a river current - water that is moving with respect to an observer on dry land. In such instances as this, the magnitude of the velocity of the moving object (whether it be a plane or a motorboat) with respect to the observer on land will not be the same as the speedometer reading of the vehicle. Motion is relative to the observer. A tailwind is merely a wind that approaches the plane from behind, thus increasing its resulting velocity. The resulting velocity of the plane is the vector sum of the two individual velocities. 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 resultant velocity of the motorboat can be determined in the same manner as was done for the plane. Motorboat problems such as these are typically accompanied by three separate questions: a) What is the resultant velocity (both magnitude and direction) of the boat? B)If the width of the river is //X// meters wide, then how much time does it take the boat to travel shore to shore? C) 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. 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. It is often said that "perpendicular components of motion are independent of each other." As applied to riverboat problems, this would mean that an across-the-river variable would be independent of a downstream variable. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">__Lesson H__ <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">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. AThe vector sum of these two components is always equal to the force at the given angle. 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.
 * 1) <span style="font-family: Arial,Helvetica,sans-serif;">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) <span style="font-family: Arial,Helvetica,sans-serif;">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) <span style="font-family: Arial,Helvetica,sans-serif;">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) <span style="font-family: Arial,Helvetica,sans-serif;">Repeat steps 2 and 3 for all vectors that are to be added
 * 5) <span style="font-family: Arial,Helvetica,sans-serif;">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) <span style="font-family: Arial,Helvetica,sans-serif;">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) <span style="font-family: Arial,Helvetica,sans-serif;">Measure the direction of the resultant using the counterclockwise convention discussed [|earlier in this lesson].
 * 1) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Select a scale and accurately draw the vector to scale in the indicated direction.
 * 2) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">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) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">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) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">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) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">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) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Construct a //rough// sketch (no scale needed) of the vector in the indicated direction. Label its magnitude and the angle that it makes with the horizontal.
 * 2) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">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) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Draw the components of the vector. The components are the //sides// of the rectangle. The tail of each component begins at the tail of the vector and stretches along the axes to the nearest corner of the rectangle. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
 * 4) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">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) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">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) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Repeat the above step using the cosine function to determine the length of the side adjacent to the indicated angle.

=<span style="font-family: Arial,Helvetica,sans-serif;">Method 3: Lesson 2 A and B = <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">Lesson A <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">1. What is a projectile? <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">- A projectile is an object that has only gravity acting upon it; if there are other forces acting upon it, it is not a projectile. <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">2. How many forces act upon projectiles? <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">- One; gravity <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">3. What are some examples of projectiles? <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif; line-height: 17px;">- An object dropped from rest without air resistance - An object thrown upward vertically without air resistance <span style="font-family: Arial,Helvetica,sans-serif;"> 4. Is a force required to keep an object in motion? Explain. <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">- No, there doesn’t have to be a force to keep an object in motion. A force is only needed to keep an object accelerating, or it would move at constant speed. <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">5. What would happen according to Newton’s first law if there were no gravity? <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">- The object would move in a straight horizontal line at a constant speed, yet with gravity it would increasingly move downward. This is due to inertia. <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">**Central Idea/Theme**: A projectile is an object with one force acting upon it: gravity. If there was no gravity, objects would continue motion at constant velocity because forces are only required for acceleration. <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">Lesson B <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">1. List two components of projectile motion. <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">- Horizontal and vertical motion <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">2. What is the acceleration of gravity? <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">- Freefalling objects fall at an acceleration of 9.8 m/s <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">3. Must there be a horizontal force for horizontal acceleration? <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">- Yes; gravity acts perpendicular to the horizontal motion although perpendicular components of motion are independent, so object moves with constant horizontal velocity at downward acceleration. <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">4. How would an object travel in constant motion at constant speed? <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">- If there was no unbalanced force, or absence of gravity; object would travel in parabolic trajectory because the downward force of gravity accelerates them downward from an otherwise straight-line, gravity-free trajectory. <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">5. How would a projectile cause a projectile to move? <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">- Gravity causes the object to move downward vertically. Gravity doesn't affect horizontal displacement. <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;"><span style="font-family: Arial,Helvetica,sans-serif;">**Central Idea/Theme**: Gravtiy does not affect the horizontal displacement of a projectile, only the vertical. =Activity: Vector Mapping= Group: Julia, Amanda, Ryan, Ben

<span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 90%; text-align: left;">
 * Conclusion**: This activity allowed us to compare the way that we find results through calculations to the actual experimenting itself. We found the percent error between what we should have got analytically and what we should have gotten graphically. We did find that the analytical percent error was larger, being 1.09%, versus the graphical percent error, which was 0.59%. This activity allowed us to see that all of the calculations that we have been doing do actually work when done physically, in real life. Some places of error could have been that the tape measurer wasn't completely straight, giving off slightly wrong values. Another source of error could be that we can only round to the nearest hundredth when measuring, while more precise and sturdy technology could be used.

=Method 3: Lesson 2 C= Part A and B 1. What would a vector diagram look like showing a projectile? It would show the x and y velocity vectors and their magnitudes labeled. Lengths of the vector arrows are representative of the magnitudes. 2. What happens to the horizontal velocity as the projectile moves? Nothing happens to it because it stays constant throughout the trajectory, moving in inertia. 3. What happens to the vertical velocity vector? It changes constantly, accelerating 9.8 m/s/s. 4. How would a projectile's x and y vectors change if it was launched horizontally? The object would move at first with initial velocity until reaching the maximum height and then accelerate once again when going down, although the horizontal will stay constant. 5. What would a v-t graph show for this event? 6. What equation can be used to show vertical displacement of a projectile? This can be shown through the formula of a free-falling object, which is y=0.5 x g x t2. 7. What do these variables stand for? G is -9.8 m/s/s and t is time in seconds; this equation refers to object with no initial velocity. 8. How could horizontal displacement of a projectile be found? Through the formula x=Vix x t. 9. What is the equation for vertical displacement for an angled-launched projectile? y=Viy x t + 0.5 x g x t2. Viy is the initial vertical velocity in m/s, t is time in seconds, and g=-9.8 m/s/s What is the position of a projectile launched at an angle to the horizontal?
 * Central Idea/Theme**: The velocity of the verticle accelerates at 9.8 m/s/s although the horizontal moves at constant speed. These circumstances are different though, in the case that gravity is present.

=Activity: Ball in Cup Part I= - Measure the initial velocity of a ball - Apply concepts from 2-D kinematics to predict the impact point of a ball in projectile motion - Take into account trial-to-trial variations in the velocity measurement when calculating the impact point media type="file" key="Movie on 2011-10-24 at 08.58.mov" width="300" height="300" media type="file" key="New Project 1.m4v" width="300" height="300"
 * Group:** Julia, Amanda, Ryan, Ben
 * Objectives**:
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">Pre Lab Questions **
 * 1) <span style="color: #000000; font-family: Arial,Helvetica,sans-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?
 * 2) I would need to know the height from the place the object fell to the ground. You must make the assumptions that there is a constant vertical acceleration of -9.8 m/s/s due to gravity, and that the initial velocity of the object is 0 m/s/s because before it was dropped, it was at rest.
 * 3) <span style="color: #000000; font-family: Arial,Helvetica,sans-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.
 * 4) <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">In all projectiles, in order for it to be considered a projectile, the horizontal velocity stays the same the entire time. Therefore, you can use this information to calculate how far it will travel before it hits the ground by plugging the horizontal velocity into distance=velocity/time, as well as the time.
 * 5) <span style="color: #000000; font-family: Arial,Helvetica,sans-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?
 * 6) <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">In order to find the time interval, you would need to find the distance that the object traveled on the x-axis (horizontally), using distance and time to find the velocity.
 * 7) <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">Write your procedure and get approval from Ms. Burns before you proceed any further!
 * 8) What data will you need to collect? Remember that you must run multiple trials. Keep in mind your end goal!
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">We need to find out what the height that the projectile will be dropping from is.
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">How will you analyze your results in terms of precision and/or in terms of accuracy?
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">Because we are able to calculate so many different aspects of the projectile through different formulas specific to them, we are able to find theoretical values. We can then use these theoretical values and compare them to experimental values that we find through our experimenting. We can use the percent error formula, and the percent difference formula to analyze how accurate our results are.
 * Hypothesis:** If our calculations are correct and we place the cup in the correct place, theoretically, then the ball should land in the cup.
 * Available Materials:**
 * Data Studio || meter stick or metric measuring tape ||
 * plumb bob || target ||
 * ramp || Carbon paper ||
 * masking tape || two right-angle clamps ||
 * Yellow Ball || newsprint ||
 * 1Photogate Timer || Calipers ||
 * Procedure:** The two videos below show how we fired the ball, aiming it at the angle of 25 degrees so that it would go into the cup.
 * Calculations:**


 * Conclusion**: This activity was a sort of preview for the Shoot for Your Grade Lab. We were able to do all of the calculations to find different components necessary to getting the ball into the cup. Things that could have accounted for some error throughout the lab was the fact that at first, we didn't account for the height of the cup. When we did recalculate though, our percent error decreased even more than it from what it had been before. Other sources of error could have been from the fact that the ball does land in a slightly different place each time, as we saw through its placement on the paper in a triangular series of dots, never landing in the exact same place. Ways to fix this error would be to have a machine which shot the ball more precisely every time. Some error is eliminated because the cup does give some leeway as to where the ball lands exactly because its radius is larger than the ball. We were, after all, able to get the ball into the cup, and this shows that if we work hard to place the cup at the right placement, we should be able to get the ball into the cup every time, or close to it.

=Activity: Ball in Cup Part 2= media type="file" key="Shoot for Your Grade.mov" width="300" height="300" Percent Error:
 * Group**: Julia, Amanda, Ryan, Ben
 * Objective:** We want to place the hoops correctly so that the ball will go through all five of them and land into the cup, using theoretical calculations as a guide.
 * Hypothesis:** If our calculations are correct and we place all of the hoops in the correct places, theoretically, then the ball should go through all 5 hoops and land in the cup.
 * Available Materials:**
 * Data Studio || meter stick or metric measuring tape ||
 * plumb bob || target ||
 * ramp || Carbon paper ||
 * masking tape || two right-angle clamps ||
 * Yellow Ball || newsprint ||
 * 1Photogate Timer || Calipers ||
 * Procedure**: We put the ball into the medium speed of the shooter, set the shooter to 25 degrees, and shot the ball. We had previously hung the hoops on the ceiling by connecting them through the ceiling tiles and securing them with tape and calipers. We placed each hoop and the cup at specific coordinates, which we found through our calculations.
 * Calculations:**
 * Conclusion**: We did calculate certain heights and distances at which we believed should be the center points of the hoops so that the ball would go through. These points included: (0.5, 0.168), (1.0, 0.210), (1.5, 0.125), (2.0, -0.088), (2.5, -0.433), and (2.953, -0.92). These heights and distances were all relative to the place from which the projectile was fired from and therefore, where it should have theoretically went. There was some error, as we found through our percent error calculations. This error ranged from 0.61% to 16.99% in the vertical values of the hoops. We had to end up guessing and checking to see where the hoop could be best positioned to match the path of the ball. We were able to get it through four hoops, still showing a significant amount of error, missing one hoop and the cup. Sources of error stemmed from the constant movement of the hoops between classes, which we couldn't control due to lack of space. Not only did we have to work with different approximations of placement, the tape that we used probably did give a little, slightly changing the position of the hoop. Due to some inconsistencies in the shooter and the angle, this could have contributed to error. We even saw on the paper how there were different distances that the ball traveled each time. This was probably caused by the angle being slightly differentiated each time. Different air pressures could have also slightly affected our results, considering that our calculations failed to account for them. We also were near an air vent, which caused some of our hoops to swing during our trials, slightly throwing off results. In order to decrease the amount of error, all groups should have had stationary individual places. Instead of using only tape, we should have use calipers to decrease movement of the strings and their give. Another way to decrease error would be to have a way of making sure that the shooter was at exactly 25 degrees each time, so that results wouldn't be affected by a changed angle. If we could measure air pressure with an instrument like a barometer, we could take it into account in our calculations to decrease percent error as well. It would also be helpful to stay in an area without any kind of ventilation nearby.

=Gordorama Contest=
 * Partner**: Amanda Fava
 * Vehicle Mass (g)**: 1.5 kg
 * Time (s)**: 8.16 s
 * Distance Traveled (m)**: 11.0 m
 * Acceleration Value**: -0.33 m/s 2
 * Acceleration Calculation and Velocity Value**:
 * Pictures**:


 * Improvements**: It would have made our cart go faster if we were able to get the axels more aligned. Our car was constructed very well, although after the first time it crashed into the wall, the axels got slightly misaligned, and therefore, the car went into the side wall during the trials, rather than going straight. I think that was definitely the main problem because our wheels seemed to be working well and avoided as much friction as possible that would slow the vehicle down and cause it to eventually stop.