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14, February 2007
Physics 124-021
Alexander Hyatt
Abstract
The purpose of this lab was to determine the initial velocity of a ball shot from a projectile launcher through several different methods, and to determine the percent difference those experimental results for initial velocity. Three different experiments were carried out with the projectile launcher as well as a ball and a meter stick to determine distances traveled. Different equations using differing variables were used to find the initial velocity from the experimental results. The three different experiments were meant to find values for the variables to be plugged into the corresponding equation to give the theoretical initial velocity. Thus, the findings of this lab were supposed to show that each of the different equations to finding the initial velocity, are correct, and that one is able to find the initial velocity based on a number of ways, depending on the variables that are known.
February 14, 2007
Physics 124-021
Alexander Hyatt
Experiment Table 1/Exercise 1
Exercise 1 Data Table Vertical displacement, y Average horizontal displacement, x Calculated time of flight, t= sqrt(2y/g) Calculated average initial velocity, vi= x/t .765 m 1.265 m .395 s 3.203 m/s
This is a table for the first exercise carried out. In this exercise, the launcher was set at an angle of zero degrees. The launcher was also set to short range setting in this exercise as well as the other two, therefore this does not make a difference in any of the results. The vertical displacement was the distance from the floor to the cross hairs of the launcher, which was measured out to be .765 meters. The average horizontal displacement was the median value in the range from the shortest shot from the launcher to the longest shot, which was measured out to be 1.265 meters. The time was calculated by using the equation, t = sqrt(2y/g). This time was calculated to be .395 sec. Finally, the average initial velocity was calculated by the equation, vi = x/t. The average initial velocity was found to be 3.203 meters per second. This average velocity will be used to compare against both of the other two excercises. This is because this value is most likely to be correct since it is the easiest to calculate, it does not rely on angles and it is the most free from human error or other types of errors.
Table 2/Exercise 2
Exercise 2 Data Table y (m) 0.012 0.02 0.047 0.082 0.124 0.181 x (m) 0.12 0.22 0.32 0.42 0.52 0.62 x2 (m2) 0.0144 0.0484 0.1024 0.1764 0.2704 0.3844
Table 2 is a table that contains the experimental data from exercise 2. The launcher was set at an angle of zero degrees. The shot was fired at a piece of paper taped to a vertical bored. An initial impact mark is on the paper is marked as the x=0,y=0 point, and this point is made by putting the vertical board with the paper up to the launcher at point blank range, so that the paper is touching the launcher. Each of the vertical distances, y, are measured from the impact mark by measuring down from the initial impact mark to the newly created impact mark. The horizontal distance, x, is measured
February 14, 2007
Physics 124-021
Alexander Hyatt
from the end of the launcher to the new distance that the board is moved for each successive trial. The initial velocity using the given data can be found by the equation, vi = sqrt(g*x2/2y). The g value is known to be the acceleration due to gravity which is 9.8 m/s2. The x2 value is just the measured x distance squared. The x2/y value can be found by graphing the x2 values on the ordinate and y values on the abscissa and finding a linear trendline. The slope of this line is this value and can be plugged into the equation. Plugging all of these values into the equation, we find that the initial velocity is found to be 3.249 meters per seconds. The percent difference between this initial velocity and the initial velocity from the first exercise is found to be 1.43%, which shows pretty good similarity between the two values.
Table 3/Exercise 3
Exercise 3 Data Table () 20 25 30 35 40 45 50 55 60 65 70 sin (2*) 0.643 0.766 0.866 0.94 0.985 1 0.985 0.94 0.866 0.766 0.643 R (m) 1.575 1.622 1.632 1.612 1.58 1.506 1.42 1.292 1.137 1.019 0.875
This is the data table for the final exercise performed. In this exercise, the launcher was set at angles, from twenty to seventy degrees. The range, R, was measured from as the horizontal difference from launcher to the impact point. The sin(2* ) value was calculated because it is used in the equation for finding the initial velocity. The initial velocity can be found, for this data, by using the equation, vi = sqrt(R*g/sin(2* )). The g value is known to be the acceleration due to gravity which is 9.8 m/s2. The R/sin(2* ) value can be found by plotting the values from the table, with the sin(2* ) values on the abscissa and the R on values the ordinate, then finding a linear trendline. The slope of this line is this value and then can be plugged into the initial velocity equation. The initial velocity is calculated out to be 2.780 meters per second. The percent difference between this initial velocity and the first exercise's initial velocity value is found to be 14.14%. This is pretty high, and can be attributed to the error with the launchers as well as the graph for finding the R/sin(2* ) value. The launcher had some errors because it maxed out its range at 30 degrees, when the maximum range should be at 45 degrees theortetically. Also, the graph used to find R/sin(2* ) had some problems, but this will be discussed in the graphs section.
February 14, 2007
Physics 124-021
Alexander Hyatt
Graphs Graph 1
Vertical Distance vs. Squared Value of Horizontal Distance
0.45
0.4
y = 2.1538x - 0.0012 R2 = 0.9981
Horizontal Distance Squared (x^2) [m^2]
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 Vertical Distance (y) [m]
This is the graph used to determine the x2/y value for the second exercise. The R2 value is very high, showing that the data very clearly resembles a linear path. The slope was found to be 2.154. This value was put into the initial velocity equation used in exercise 2, and the initial velocity was found to be 3.249 meters per second.
February 14, 2007
Physics 124-021
Alexander Hyatt
Graph 2
The Sin of 2 times Theta vs. the Range of the Shot
1.8
1.6 y = 0.7884x + 0.7145 R2 = 0.1507 1.4
1.2
Range (R) [m]
1
0.8
0.6
0.4
0.2
0 0 0.2 0.4 0.6 Sin of 2 Theta (sin(2*theta)) [deg] 0.8 1 1.2
This is the graph used to find the R/sin(2* ) value for the third exercise. The R2 value is very off, because the data is very scattered at the beginning of the graph. The graph was saved perhaps because the data starts to get closer and closer together as the graph's x value increases. Plugging in this value into the initial velocity equation used in exercise 3 gives a value of 2.780 meters per second. The inaccuracy of the data is not as bad as the R2 value because the percent difference between the initial velocity for exercise 3 and exercise 1 is 14.14%. While this is not great, it is not incredibly terrible either. Most of the error can be attributed to the error with the launcher as described above. Also, the graphs trendline was an error that attributed to this fairly high percent difference value.
February 14, 2007
Physics 124-021
Alexander Hyatt
Questions Calculation Details
Exercise 1 Time = sqrt(2*y/g) = sqrt(2*.765/9.8) = .395s Initial velocity = vi = x/t = 1.295m/.395s = 3.203 m/s Exercise 2 Initial velocity = vi = sqrt(g*x2/2*y) = sqrt(9.8m/s2 *2.154m/2) = 3.249m/s Exercise 3 Initial velocity = vi = sqrt(g* R/sin(2* )) = sqrt(9.8m/s2*.7884m) = 2.780m/s Percent Difference = |measured2 measured1|/((measured1 + measured2)/2)*100 = |3.249 3.203|/((3.203 + 3.249)/2)*100 = 1.43%
Questions
1. Do the Results show that the projectile launcher is both an accurate and precise piece of equipment? The results show that the projectile launcher is not a very accurate or precise piece of equipment. This is mostly because the initial velocity value found in exercise 3 is so different than that of both exercise 1 and 2. This shows that there are probably problems with the angles that the launcher is set at. The launcher may need to be calibrated again. 2. From the kinematics equations (3.1-3.4), derive equiation 3.10, which shows that when an object is dropped or launched horizontally at a height, y, the time to reach the ground is t = sqrt(2*y/g). *g*t2 = y => g*t2 = 2*y => t2 = 2*y/g => t = sqrt(2*y/g) 3. Derive equation 3.11. x2 = vi2*t2 y = *g*t2 => vi2*y = *g*vi2*t2 => 2*vi2*y = g*x2 = y = g*x2/2*vi2
February 14, 2007
Physics 124-021
Alexander Hyatt
4. Use equation 3.6 to show that if an object is launched from the ground, the total time it is in the air is given by t = (2*vi*sin()/g). vi*sin()-g*t => g*t = vi*sin() => t = vi*sin()/g time up = time down so, t = (2*vi*sin()/g) 5. Using the results from exercise 3, what can be said for projectiles launched at complimentary angles? Complimentary angles are those that add to 90 deg, e.g., 60 deg and 30. The lower angle's range value divided by the upper angle's range value gets closer and closer to one as both angles approach 45 degrees. This shows that the maximum range is or close to 45 degrees. 6. Use equation 3.12 to determine the angle at which a maximum range is obtained, given one consistent launch velocity. Do your exercise 3 results verify this? Explain why or why not. R = sin(2*)/(9.8) (graph) Using a graphing calculator, the max was to be determined at 44.9995 deg. which rounds off to 45 degrees. This is not consistent with my data because my largest range was at about 30 degrees.

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