48.0-kg projectile is fired at an angle of 30.0° above the horizontal with an in

Question

48.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 1.28 102 m/s from the top of a cliff 154 m above level ground, where the ground is taken to be y = 0.
(a) What is the initial total mechanical energy of the projectile?
__________ J

(b) Suppose the projectile is traveling 90.7 m/s at its maximum height of y = 325 m. How much work has been done on the projectile by air friction?___ J

(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?
____ m/s

Explanation / Answer

a) total mechanical energy = KE + PE = (1/2)*mv2 + mgh = 465731.52J

b) Find total mechanical energy (KE + PE) at max height. Subtract this from the part a) answer.
This give the work done by friction (W) as some of the mechanical energy has been 'lost' due to friction.

465731.52 - (0.5*mv2 + mgh) = 115259.76J

c) Energy lost due to friction going up =W; energy lost coming down =1.5W. Total = 2.5W.
Subtract 2.5W from the part a) answer. This is the mechanical energy as it hits the ground. Since y=0, the energy is all KE. So find v using KE = (1/2)*mv2

465731.52 - 2.5W = 177582.12J = (1/2)*mv2 ----> v = 86 m/s

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