A cannon, located 43.0m from the base of a vertical 26.6m tall cliff, shoots a 1

Question

A cannon, located 43.0m from the base of a vertical 26.6m tall cliff, shoots a 15.0kg shell at 41.4° 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 26.6m 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

consider the motion along the X-direction

Vox = initial velocity = Vo Cos41.4

t = time of travel

X = displacement = 43 m

using the equation

X = Vox t

43 = ( Vo Cos41.4) t

t = 43 / ( Vo Cos41.4) eq-1

consider the motion along the Y-direction

Voy = initial velocity = Vo Sin41.4

t = time of travel

Y = displacement = 26.6 m

using the equation

Y = Voy t + (0.5) a t2

26.6 = (Vo Sin41.4) (43 / ( Vo Cos41.4) ) + (0.5) (- 9.8) (43 / ( Vo Cos41.4) )2

Vo = 199.56 m/s

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