A 60.0kg projectile is fired at 30.0 degree above the horizon at an intial spped

Question

A 60.0kg projectile is fired at 30.0 degree above the horizon at an intial spped of 124 m/s from the top of the cliff 114m above level ground, where the ground is taken to be y=0 a) what is the intial total mechanism energy of the projectile? b) Suppose the projectile is traveling 87.9 m/s at its maximum height of y = 274 m. How much work has been done on the projectile by air friction? c)What is the speed of the projectile immediately before it hits the ground if air friction does one and a half times as much work on the projectile when it is going down as it did when it was going up?

Explanation / Answer

Part A)
Total enegry = KE + PE

E = (.5mv^2) + (mgh)

E = (.5)(60)(124)^2 + (60)(9.8)(114)

E = 528312 J

Part B)

E = (.5mv^2) + (mgh)

E = (.5)(60)(87.9)^2 + (60)(9.8)(274)

E = 392904.3 J

The work by friction is 528312 - 392904.3 = 135408 J

Part C)

The loss will be 135408(1.5) = 203112 J

The KE will be 392904.3 - 203112 = 189793 J

KE = .5mv^2

189793 = (.5)(60)(v^2)

v = 79.5 m/s

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