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Initial Velocity of a Projectile

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Lab #3: Initial Velocity of a Projectile| | | Abhishek Samdaria| Pd. 4 and 5| | Lab #3: Initial Velocity of a Projectile Theory: How can we determine the initial velocity of a projectile? Experimental Design: The purpose behind this experiment was to determine the initial velocity of a projectile. Projection motion consists of kinematics of motion in the x and y directions. With two dimension kinematics, there are the x and y components in any given velocity. In projectile motion, the x component has no acceleration as no outside forces are acting on it.

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The Y component on the other hand has gravity acting as a force. A small ball is shot, at three various angles (30,45,60), and through the known values the initial velocity of the ball is found. As a result, the range of the project can be represented with the equation 1) R = V02g*Sin2? , where R represents the range or Dx; the values of g and ? are known. However, in this experiment, one main equation were used to determine the initial velocity.

1) y-y0=tan? x-gx22(V0cos? )2 , where y is the trajectory of a particle in two dimensional motion, gravity is -9. 1 m/s 2 , and ? is the launch angle. X is equal to the average distance launched in the x direction. In order to determine all the components required to use the trajectory equation, a small projectile ball was launched at 3 different angles. The distance traveled was measured as well as the initial height the ball was launched. 3 trials of each angle were conducted and the average of the trials was used as the x distance in order to determine the trajectory. The photo-gate was used to find how long it took the diameter of the ball to pass through the sensor.

Using all the data gathered, the initial velocity should be able to be determine as it is the only missing variable. Materials and Methods The materials needed were the projectile launcher, the plunger, a spherical ball, a photogate, a meter stick and white sheets of paper. To go about the experiment, first the angle of the projectile launcher to the horizontal was set to 30 degrees and the height from the ground to the bottom of the launching position was measured. Then a trial shot was fired to approximate the location of the end of the ball’s trajectory.

Then the white papers were placed in the area around the location from the first shot. Trial one was initiated and the ball was shot again by pushing it into the launcher with a plunger like object and then pulling the cord. The landing spot of the ball was recorded on the paper on the ground and the distance from the point where the ball hit the ground to the location of the launcher. The time it took for the ball to pass through the photo-gate was recorded. Then two more trials were done. Then the angle was changed to 45 degrees and the procedure was repeated.

Finally the angle was changed to 60 degrees and the procedure was repeated for a last time. Data: Table 1:Time (sec) for diameter of the ball to pass through the photo-gate ? | Trial 1| Trial 2| Trial 3| Average| V0(ms)| 30| . 0081| . 0079| . 0082| . 008067| 2. 727| 45| . 0043| . 0044| . 0065| . 005067| 4. 342| 60| . 0046| . 0044| . 0046| . 004333| 5. 077| Table 2:Distance(m) the projectile travelled ?| Trial 1| Trial 2| Trial 3| Average| 30| 2. 3| 2. 36| 2. 37| 2. 3433| 45| 2. 5| 2. 64| 2. 71| 2. 6167| 60| 2. 2| 2. 13| 2. 08| 2. 1367| This is the data of all the different trials required to calculate the initial velocity. -y0=tan? x-gx22(V0cos? )2 y=1. 03m 1. 03-y0=tan? x-gx22(V0cos? )2 y0=0 1. 03-0=tan? x-gx22(V0cos? )2 ?=30° 1. 03=tan? (30)x-gx22(V0cos? (30))2 x=2. 3433m 1. 03=tan30(2. 3433)-g(2. 3433)22(V0cos? (30))2 g=-9. 81ms2 1. 03m=tan302. 3433m-9. 81ms2 (2. 3433m)22(V0cos? (30))2 1. 03m-tan302. 3433m=-9. 81ms2 (2. 3433m)22(V0cos? (30))2 (2cos2(30))(1. 03m-tan30(2. 3433m))=9. 81ms2 (2. 3433m)2V02 -0. 481499-9. 81ms2 (2. 3433m)2=1V02 1. 09117=1V0 10. 97ms=V0 This is the data of all the different trials required to calculate the initial velocity. y-y0=tan? x-gx22(V0cos? )2 y=1. 05m 1. 05-y0=tan? -gx22(V0cos? )2 y0=0 1. 05-0=tan? x-gx22(V0cos? )2 ?=45° 1. 05=tan? (45)x-gx22(V0cos? (45))2 x=2. 6167m 1. 05=tan45(2. 6167)-g(2. 6167)22(V0cos? (45))2 g= -9. 81ms2 1. 05m=tan452. 6167m-9. 81ms2 (2. 6167m)22(V0cos? (45))2 1. 05m-tan452. 6167m=-9. 81ms2 (2. 6167m)22(V0cos? (45))2 (2cos2(45))(1. 05m-tan45(2. 6167m))=9. 81ms2 (2. 6167m)2V02 -1. 567-9. 81ms2 (2. 6167m)2=1V02 6. 55ms=V0

This is the data of all the different trials required to calculate the initial velocity. y-y0=tan? x-gx22(V0cos? )2 y=1. 07m 1. 07-y0=tan? x-gx22(V0cos? )2 y0=0 1. 07-0=tan? x-gx22(V0cos? )2 ?=60° 1. 07=tan? 60)x-gx22(V0cos? (60))2 x=2. 1367m 1. 07=tan60(2. 1367)-g(2. 1367)22(V0cos? (60))2 g= -9. 81ms2 1. 07m=tan602. 1367m-9. 81ms2 (2. 1367m)22(V0cos? (60))2 1. 07m-tan602. 1367m=-9. 81ms2 (2. 1367m)22(V0cos? (60))2 (2cos2(60))(1. 07m-tan60(2. 1367m))=9. 81ms2 (2. 1367m)2V02 5. 835ms=V0 Measured V0(ms) calculations: Diameter(of sperical ball)Time to pass through photo-gate ?: 30 degrees -. 022. 008067 = 2. 727ms ?: 45 degrees -. 022. 005067 = 4. 342ms ?: 60 degrees -. 022. 0043 = 5. 077ms Percent Error: Percent Difference Calculation: ?: 30 degrees * % Difference = measured-theorticalmeasured*100 % Difference = (2. 727ms)—(-10. 97ms)(2. 727ms)*100 * % Difference = 302. 25% ?: 45 degrees * % Difference = measured-theorticalmeasured*100 * % Difference = (4. 342ms)—(-6. 55ms)(4. 343ms)*100 * % Difference = 50. 86% ?: 60 degrees * % Difference = measured-theorticalmeasured*100 * % Difference = (5. 077ms)—(-5. 835ms)(5. 077ms)*100 * % Difference = 14. 93% Uncertainty Analysis Discussion There were measured quantities with uncertainty values in this experiment. It was impossible to tell if the ball was released at either the exact defined height or a bit above it.

The meter stick measured by centimeters so the uncertainty value was half a centimeter. It was also impossible to tell the definite diameter of the ball. The caliper measured by millimeters so the uncertainty was half a millimeter. The most dominant source of uncertainty was the range that the ball travelled and the initial height the ball was launched from. Conclusion From the data gathered we can conclude that the average initial velocity of the ball could be anywhere from 4 to 6 meters per second. The accepted range of times for the complete diameter of the ball to pass the photo-gate is . 004sec -. 005sec. Evaluate:

There are sources of error that could have accounted for the percent error seen. The first source of error is air resistance. The actual value is determined in a vacuum setting where there is no air friction. The second potential source of error is the fact that the projectile launcher was not permanently positioned on the table and moved after every shot. That means that the direction in which the ball travelled was not always constant. This experiment was overall helpful in understanding the uses of the trajectory equation and how it can be used in order to calculate the initial velocity without the use of time.

In order to improve the experiment, the ball should be shot in a vacuum in order to minimize the effects of air resistance on the projectile when it is launched. Improvements? There can be a few improvements on the experiments. The data can become unskewed by performing the experiment in a vacuum, using a machine that would keep the launcher positioned in the same place and orientations when the trigger is pulled, and using a machine that would allow measuring the distances to a more precise degree. These improvements, however, may cost a lot of resources and money, so the experiment now is suffice.

Cite this Initial Velocity of a Projectile

Initial Velocity of a Projectile. (2017, Jan 31). Retrieved from https://graduateway.com/initial-velocity-of-a-projectile/

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