A 48.0-kg projectile is fired at an angle of 30.0 degree above the horizontal wi

Question

A 48.0-kg projectile is fired at an angle of 30.0 degree above the horizontal with an initial speed of 1.38 x 102 m/s from the top of a cliff 146 m above level ground, where the ground is taken to be y = 0. What is the initial total mechanical energy of the projectile? Suppose the projectile is traveling 97.8 m/s at its maximum height of y = 344 m. How much work has been done on the projectile by air friction? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. 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

a)
=.5mv^2+mgh
=.5*48*1.38e2^2+48*9.8*146=5.25e5 J

b)
.5*48*97.8^2+48*9.8*344=3.9137e5 J

so the work done = 5.25e5-3.9137e5
=1.3363e5 J

c)
total work done by air=(1+1.5)*1.3363e5
=3.34075e5

at ground mgh=0
so .5*48*V^2+3.34075e5=5.25e5
V^2=7955.208
V=89.19 m/s

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