A cannon, located 47.4m from the base of a vertical 28.8m tall cliff, shoots a 1

Question

A cannon, located 47.4m from the base of a vertical 28.8m tall cliff, shoots a 15.0kg shell at 41.4° above the horizontal toward the cliff

1. What must the minimum muzzle velocity be for the shell to clear the cliff edge?

2. The ground at the top of the cliff is level, with a constant elevation of 28.8m above the cannon. How far past the cliff edge does the shell land when the shell is fired at that minimum speed?

3.What is the magnitude of the acceleration of the shell just before it hits the ground?

4. 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?

Please answer all questions.

Explanation / Answer

height of cliff h = 28.8 m
distaqnce of canon form cliff, d = 47.4 m
mass of shell = m =15 kg
angle of projectile theta = 41.4 deg

a. minimum muZZle velocity to clear the cliff be v
   then at time t
   d = vcos(theta)*t
   and h = vsin(theta)*t - 0.5gt^2
   h = vsin(theta)*d/vcos(theta) - 0.5*g*d^2/v^2*cos^2(theta)
   h = tan(41.4)*47.4 - 0.5*g*47.4^2/v^2*cos^2(41.4)
   28.8 = 41.7887 - 19585.945/v^2
   v = 38.83 m/s

b. using equaiton of projectile
   y = xtan(theta) - x^2/v^2cos^2(theta)
   x^2/v^2cos^2(theta) - x*tan(theta) + y = 0
   now, y = 28.8 m
   so, 0.0011786x^2 - 0.8816x + 28.8 = 0
   solvinf for x
   x = 713.78m
   so the canon ball travels addiitonal 713.78 - 47.4 = 666.38 m on the surface of the cliff

c. magnitude of acceleration of the shell, just before it hits the ground = g = 9.1 m/s/s directed downwards
d. for this launch angle, the canon can be as far as713.78 m from the cliff

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