A cannon, located 44.8m from the base of a vertical 26.4m tall cliff, shoots a 1
ID: 1326103 • Letter: A
Question
A cannon, located 44.8m from the base of a vertical 26.4m tall cliff, shoots a 15.0kg shell at 43.0° above the horizontal toward the cliff.
What must the minimum muzzle velocity be for the shell to clear the cliff edge?
The ground at the top of the cliff is level, with a constant elevation of 26.4m above the cannon. How far past the cliff edge does the shell land when the shell is fired at that minimum speed?
What is the magnitude of the acceleration of the shell just before it hits the ground?
For this launch angle, what is the furthest distance to the cliff's base that the cannon can be moved to and still reach the level ground at the cliff's top given this launch angle and velocity?
Explanation / Answer
1.
As
R = vo^2 sin(2*angle) / g
Plugging in R = 44.8 m, angle = 43 degrees, g = 9.8 m/s^2, and solving for vo,
vo = 20.979 m/s [ANSWER]
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2.
The components of this velocity are
vox = 15.343 m/s
voy = 14.308 m/s
The time it takes to fall by 26.4 m can be obtained by using
delta y = voy t - 1/2 g t^2
as delta y = -26.4 m,
t = 4.2021 s
Thus, the bullet travels a total of
delta x = 15.343 m/s * 4.2021 s = 922.4 m/s
Thus, this is
d = 922.4 - 44.8 m= 877.6 m from the cliff. [ANSWER]
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3.
The acceleration is always g = 9.8 m/s^2 [ANSWER]
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4.
Please check the phrasing of this part.
We have fitted the problem so that the bullet barely clears the cliff, so the answer here must be 44.8 m. [ANSWER]
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