Ani,+Sammy,+Ariel+Projectile+Project

=Shoot Your Grade Lab=

A parabolic trajectory is the path of motion of a projectile. A projectile is only acted upon by one force – gravity. It consists of 2-dimensional motion with an x and y component. The projectile’s path can be illustrated by a parabolic graph. Horizontal acceleration always equals 0, because there are no horizontal forces acting on the object so the object remains at a constant horizontal velocity. The vertical acceleration is always -9.8 m/s/s, the acceleration of gravity. In this project, our goal was to observe the trajectory of a projectile and determine the affect of different angles on the path of the ball. Our ultimate goal was to calculate the ideal distance and projection angle of the ball to go through a target at any random height. By testing different angles and measuring the horizontal and vertical distances traveled for each, we calculated the initial velocity of the ball being projected by the launcher. We chose to test common angles with trigonometric identities: 0, 30, 45, and 60 degree angles. We found the average horizontal and vertical distances traveled for the ball at each angle and used the projectile equations to determine the total time of the projection and eventually the initial velocity of the ball. With this information, we were prepared to determine the specific projection angle needed to shoot the ball through the randomly placed hoop at the ball’s maximum height. On an excel spreadsheet, we used equations in the cells to, given a maximum height needed, extract the correct angle and horizontal distance from the target.

Procedure
The data shows the angle, horizontal distance, vertical distance, acceleration, time, initial velocity, initial velocity x, initial velocity y, time to maximum height, y-distance to maximum height, and x-distance to maximum height. The angles chosen for calibration are based on the trigonometric identities. The horizontal distance (x), was measured using a tape measure. For vertical distance (y), a distance of 0.28 meters was a constant, as that was always the distance the ball was from the ground in the launcher. It is known that y acceleration equals -9.8 m/s/s and x acceleration equals 0.0 m/s/s. For each of the angles, time was calculated using the equation : dy = visinØt + ½ at^2 once initial velocity was known. To calculate initial velocity, the equation, dx = vicosØt was used. After knowing initial velocity, the x initial velocity was found by multiplying vi by cosØ, and the y velocity was found by multiply vi by sinØ. After knowing all of this information and knowing that final velocity equals zero, the time to maximum height was calculated using the equation Viy + ayt. The time found was used in the distance equations, along with initial velocity, and acceleration, to calculate the x and y distances to maximum height. For each of these calculations, the average result was taken into account to use for performance day calculations. The performance day calculations took into account the above information because initial velocity was needed. Information from the calibration helps us make sure the performance day information is correct and in context. It allowed us to see what angle would provide the maximum height what distance we should put the launcher at from the maximum height. Based on the calibration data, the ball reached maximum height of 1.42 meters at around 45˚ at an x-distance away of 2.28 meters.

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The materials we used for the trials included a projectile launcher, a baton to push the projectile into the launcher, a small yellow ball (projectile), and 2 meter sticks. We used a roll of Scotch tape as a target through which we practiced launching the ball.======

Graph 1: We constructed this graph on excel using our horizontal distance traveled values and angle values based on our trials using the projectile launcher. The purpose of this graph is to extrapolate or interpolate information about the theoretical horizontal distance traveled given any angle. When we fitted a regression line, it was parabolic, and had a very high r^2 value, indicating a good fit trustworthy and precise results. We used this graph to determine if our calculations for x-distance given a max height were legitimate and logical.

Graph 2: We constructed this graph on excel using our horizontal distance traveled values for each angle based on our trials using the projectile launcher and our calculated time values for the relative angle. The purpose of this graph is to extrapolate or interpolate information about the theoretical horizontal distance traveled given any specific total time for an angle. We used this graph to help us on presentation day by comparing the theoretical time on this graph with the calculations we used to figure out time given max height. We wanted to see if our calculations gave us realistic and logical results. When we fitted a regression line, it was parabolic, and had a very high r^2 value, indicating a good fit and trustworthy and precise results.

Graph 3: We constructed this graph also on excel using our angles used in our trials and our calculated values of time. The purpose of this graph is compare the travel time based on a given angle. We used this graph by comparing the time given a projection angle to the time given the horizontal distance traveled by that same angle. This graph was helpful only in conjunction with the other two, allowing us to look at all of the pieces of our calculations at once compared to all of the pieces of our trials. A linear pattern is illustrated by this graph, implying a direct positive correlation between a projection angle and the amount of time the projectile spends in motion. When we fitted a regression line, it was linear, and had a very high r^2 value, indicating a good fit and trustworthy and precise results.

Calculations (see sample calculation sheet: [[file:Ani calculations.pdf]])
Initial Velocity: To find the initial velocity for each angle, we used the following equations: 0˚: dy=visinØt+0.5at^2 and dx=vicosØt For the angle 0˚, when finding dy, we were able to use the simplified equation dy = 0.5at^2 because sin (0)=0. All other angles: using dy=visinØt+0.5at^2 and dx=vicosØt to derive the equation: dy=dxtanØ+0.5at^2 We combined the equations (see example calculation sheet) using our given values of acceleration and our observed values of, dx and dy to calculate vi and t for the launcher at each angle.

Maximum Height: to find time to maximum height: Vfy=Viy+ayt, where Vfy is 0, because the projectile is not moving up or down at maximum height, just horizontal to find maximum height: Vfy^2=Viy^2+2aydy distance to maximum height: dx=Vixt, where dx is horizontal distance to maximum height and t is time to maximum height

Given Maximum Height: To find the angle and horizontal distance needed to reach maximum height on excel, we used the same kinematics equations as above. However, we first solved for VisinØ^2, using that value to solve for Ø, and finally plugging in Ø to solve for time to maximum height, which we used to find the horizontal distance to maximum height. The horizontal distance and the projection angle are what we used on presentation day.

Assumptions Made
The first assumption we made is the x component is the horizontal component and the y component is the vertical component. While doing calculations, we assume first and foremost that there is no air resistance. We assume no air resistance in this lab because the projectile in this case is very small, has a small mass, and wouldn’t be affected significantly by air resistance because of the minimal travel time. Next, we assumed the vertical acceleration of the projectile was -9.8 m/s/s. This value can be assumed because gravity is the only force acting on the projectile. -9.8 m/s/s is the acceleration caused by gravity on earth. It is a common constant so this number is widely accepted and used in physical calculations. We also assumed that the horizontal acceleration was 0 m/s/s. This value can be assumed because there is no horizontal force acting on the projectile. In this lab, we are assuming there is no air resistance. 0 m/s/s for horizontal acceleration is a common constant so this number is widely accepted. Finally, we assumed that the averages of the data we recorded in our trials would be the best values to use in our calculations. The projectile launcher does not always yield 100% consistent values. By using the averages of the data we observed, we assumed our calculations would be most precise.

Procedure
Using measuring tape, distance from the ground to the middle of the hoop was measured, where the ball needed to go through. The distance found was 1.45 meters. Once this distance was known, we were able to plug it into the excel sheet, and calculate the angle and x-distance needed to get the ball through the hoop. The first equation used was: to solve for visinØ^2 It was visinØ^2 = 0- ((max height + y-distance)* 2ay)). This is from the equation vfy^2 = viy^2+2a∆d. visinØ^2 = 22.25. After calculating visinØ^2, Ø was calculated by taking the inverse sin of the square root of visinØ^2 divided by the average x velocity, which was calculated prior to launch day. So, Ø=arcsin*sqrt(visinØ^2/vix) and Ø = .784 radians. Since angles are calculated in radians on excel, the angles had to be converted into degrees for the launch. That is done by multiplying the value of the angles in radians by 180, and dividing it by 3.14. Ø = 44.9˚. The launcher’s angle was set to 44.9˚. Next, the time to maximum height is calculated by multiplying the average velocity by sin of the radians, and dividing by 9.8. We then use the time to calculate x distance, using the distance equation. dx=vicosøt The x distance was 1.18 meters away from the hoop.

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For performance day, we used a textbook as a base for the launcher, the meter sticks again to measure the horizontal distance, a roll of scotch tape acted once again as a target, and the launcher, ball, and baton.======

Observations/Data from Performance
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Our first shot didn’t quite make it through the hoop. We thought the springs just had to warm up, but the ball still did not make it through the hoop on our second shot. Since spring problems often occurred, the ball was shot at this angle again. After that didn’t work, the x-distance was changed. Unfortunately, the ball was still too low. Then, the vertical distance was adjusted by placing the launcher on the textbook. This, however, did not work either. For some reason, the ball was reaching maximum height about a meter after the hoop. We hypothesized that there was a spring problem. However, we decided to look back to our excel sheet to make sure that there wasn’t an error. Despite checking it over a lot before launch day, we realized there was a typo after launch day. Instead of using average velocity in our calculation for time to maximum height, we used average y velocity by accident. When we fixed the error, we found that the x-distance should have been one meter farther back from the hoop. The correct x-distance should have been 2.28 meters

The margins of error for the angle measurements are +/- 1 degree. For the horizontal and vertical distances measured, the margins of error are +/- 0.001 m. The accelerations are given constants, so they are very precise. After using these rounded values in the equations to determine initial velocity, the percent margin of error was 3% (see calculations below). The maximum heights should be rounded to the nearest centimeter for the most precise results. We got this percent error by finding the absolute value of the highest velocity subtracted by the average velocity all divided by the highest velocity added to the average velocity divided by 2. We used this to determine the margin of error for the maximum height.

|highest – average| / ( ½ (highest+average)) |6.88 – 6.68| / (½ (6.68+6.88)) = 0.03 --> **3% percent margin of error** Before we caught the error in our excel spreadsheet, we had different explanations for what could have went wrong in our trials and on presentation day. One major source of error could have come from the springs. When we were practicing, the same angle often produced results that were extremely different. The ball went closer and further even without changing the projectile launcher. This could have been solved a few different ways. One thing that could have been done differently would be more trials for each angle being examined. We did about 3 trials for each angle, but if this number were larger, we would have probably had a more accurate average velocity to work with. Another thing that could have yielded better results would have been to warm up the springs more times than we did. If they were warmed up closer to when we actually launched our projectile, the springs would have been better prepared for the real thing.

** Conclusions **
In conclusion, our ball missed the target all five trials. Our biggest issue was that the ball was going below the target, and reaching its maximum height after it had already passed the 5 cm ring. This problem could have been solved by changing a minor typing error in our excel spreadsheet. We calculated a small margin of error, 3%, so if our excel spreadsheet had been fixed before presentation day; the ball most likely would have made it into the ring at its appropriate maximum height. Our original calculations used the average y velocity instead of the average velocity of the projectile. This gave us a shorter x-distance. If the actual average velocity of the projectile was used however, the distance would have been further away by about 1 meter, and would have reached max height where and when it was supposed to.