cannon, located 43.6m from the base of a vertical 27.8m tall cliff, shoots a 15.

Question

cannon, located 43.6m from the base of a vertical 27.8m tall cliff, shoots a 15.0kg shell at 42.0degree 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 27.8m 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 cliffs base that the cannon can be moved to and still reach the level ground at the cliffs top given this launch angle and velocity?

Explanation / Answer

From the given data the total vertical height is given by

h =y1+y2 =43.6m+27.8m =71.4m

Now let u be the minimum muzzle velocity required for the shell to lear the top of the cliff then

h =u2sintheta/2g

71.4m =u2sin2(42)/2*9.81

Then u =Sqrt(71.4*2*9.81/sin2(42))=55.937m/s

Then time taken to fall on to the ground is given by

t =2usintheta/g =2*55.937m/s*sin(42)/9.81m/s2 =7.630s

Now the distance at which the shell land is

d =horizontal compoenent of velocity * time =55.937cos(42)*7.630 =317.173m

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