Week 4 Lab
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Projectile Motion Lab
Initial Observations: 1.
How does changing the launch angle affect the trajectory, range, or flight time?
Changing the launch angle affects all 3 of these categories during projectile motion. The object's trajectory has a direct correlation to the angle because as I've observed, the higher
the angle, the higher the trajectory and the peak of the flight, this applies to the opposite case, where we have smaller and smaller angles and thus a straighter and lower trajectory.
Regarding range, it has a cutoff for both cases. the lower we go from about 90 to 45 degrees, the range gets larger and larger, but when we reach 45 degrees and below, our range starts to shrink again. Regarding flight time, the higher the angle, the higher the flight time due to how much higher it has to go. 2. How does changing the initial speed affect the trajectory, range, or flight time?
If we keep the angle the same as well as the mass of the cannonball during launch and only change the initial speed what we find is intuitive. The more initial speed we give the object,
the trajectory will keep on getting higher, the range will keep increasing, and since both of these take longer our flight time will also proportionally increase. 3. How does changing the mass affect the trajectory, range, or flight time?
No matter how much higher or lower the mass is changed, the trajectory, range, and time all stay the same.
Explanation: I chose 5 angles that go up by 15 degrees from the previous. I had a constant mass of 29.37kg. The diameter was a constant 0.18m, and gravity was not changed from 9.81 m/s^2. Each trial was set to the exact value described, and the data was collected with the tools provided
on the website. Angles and Trials :
Initial plot of Angle vs Max Height with Constant velocity
Linearized Angle^2 vs Max Height ^2 with Constant Velocity: Angle
(Degrees)
Launch Velocity (m/s)
Max Height
(m)
±0.005m
Range
(m)
±0.005m
Flight Time
(s)
±0.005s
30
20m/s
5.090m
35.310m
2.040s
45
20m/s
10.190m
40.770m
2.880s
60
20m/s
15.290m
35.310
3.530s
75
20m/s
19.020m
20.390m
3.940s
90
20m/s
20.390m
0.000m
4.080s
Angle not Angle^2
Process:
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Brief discussion: Based on my expected value of the linearized graph found by using point slope to find the slope based on the new points given by the new linear equation, the equation matches based on the uncertainties found for both the slope and y-intercept.
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