M3
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Broward College *
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2048L
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Aerospace Engineering
Date
Apr 3, 2024
Type
docx
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7
Uploaded by SargentGalaxy13589
M3.1 Laboratory Report 3
Name: Zhiyuan Wu
Date: 1/29/2024
Propose: Use a computer simulation to investigate what projectile
motion is.
Introduction: Kinematics equations
,
gravitational acceleration
Procedure: Use a computer simulation, input angle, initial speed and
so on. Then see the vectors of acceleration.
Data and Data Evaluation:
Part 1
Pumpkin:
Initial Speed
Time
Range Height
2 m/s
1.28 s
2.55 m
0 m
4 m/s
1.28 s
5.11 m
0 m
6 m/s
1.28 s
7.66 m
0 m
8 m/s
1.28 s
10.22 m
0 m
10 m/s
1.28 s
12.77 m
0 m
12 m/s
1.28 s
15.33 m
0 m
14 m/s
1.28 s
17.88 m
0 m
16 m/s
1.28 s
20.43 m
0 m
18 m/s
1.28 s
22.99 m
0 m
20 m/s
1.28 s
25.54 m
0 m
These values show with the increase of initial speed, the range is
increasing, but the time of landing is not change.
Car:
Initial Speed
Time
Range Height
2 m/s
1.28 s
2.55 m
0 m
4 m/s
1.28 s
5.11 m
0 m
6 m/s
1.28 s
7.66 m
0 m
8 m/s
1.28 s
10.22 m
0 m
10 m/s
1.28 s
12.77 m
0 m
12 m/s
1.28 s
15.33 m
0 m
14 m/s
1.28 s
17.88 m
0 m
16 m/s
1.28 s
20.43 m
0 m
18 m/s
1.28 s
22.99 m
0 m
20 m/s
1.28 s
25.54 m
0 m
Comparing the two results, I find with the same initial speed,
whatever the object is, the time to landing and range will not
change.
Part 2
investigating gravitational acceleration
The direction of the vector is down, and it represents gravitational
acceleration.
The length of the vector doesn’t change throughout its flight.
It can tell me the direction and magnitude of the acceleration acting
on the cannonball throughout its duration of flight.
If we change the angle of the cannon, I think the acceleration vector
will not change, because the gravitational acceleration is constant.
The acceleration vector does not change its length and direction at
new angle, my prediction is right.
Discovery: The launch angle affects the trajectory and range of the
projectile but does not affect the gravitational acceleration acting on
it.
investigating velocity
Angle 45 degree:
The length of the velocity vector in the y-direction decreases as the
cannonball rises, becomes zero at the peak of its trajectory, and
then increases as the cannonball falls back down.
Initially upward, it becomes zero at the peak, and then downwards
as the cannonball descends.
The velocity vector in the y-direction becomes zero at the peak of
the trajectory. This is the highest point of the flight, where the
cannonball momentarily has no vertical motion.
Ascent: On leaving the cannon, the vertical component of velocity is
maximum and positive. It decreases due to gravity.
Peak: At the peak, all the upward velocity is counteracted by
gravity, making the vertical velocity zero.
Descent: After the peak, the cannonball gains downward velocity,
increasing in magnitude until it hits the ground.
The velocity in the x-direction remains constant throughout the
flight.
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