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. 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*Sin2θ , where R represents the range or Dx; the values of g and θ are known.
However, in this experiment, one main equation were used to determine the initial velocity. 1) yy0=tanθxgx22(V0cosθ)2 , where y is the trajectory of a particle in two dimensional motion, gravity is 9.81 m/s 2 , and θ 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 photogate 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 photogate 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 photogate θ 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
θ 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.
yy0=tanθxgx22(V0cosθ)2
y=1.03m
1.03y0=tanθxgx22(V0cosθ)2
y0=0
1.030=tanθxgx22(V0cosθ)2
θ=30°
1.03=tan(30)xgx22(V0cos(30))2
x=2.3433m
1.03=tan30(2.3433)g(2.3433)22(V0cos(30))2
g=9.81ms2
1.03m=tan302.3433m9.81ms2 (2.3433m)22(V0cos(30))2
1.03mtan302.3433m=9.81ms2 (2.3433m)22(V0cos(30))2
(2cos2(30))(1.03mtan30(2.3433m))=9.81ms2 (2.3433m)2V02
0.4814999.81ms2 (2.3433m)2=1V02
1.09117=1V0
10.97ms=V0
This is...