Question

A 44.0-kg projectile is fired at an angle of 30.0° above the
horizontal with an initial speed of 144 m/s from the top of a cliff
120 m above level ground, where the ground is taken to be
*y* = 0.

(a) What is the initial total mechanical energy of the
projectile? (Give your answer to at least three significant
figures.)

(b) Suppose the projectile is traveling 102.1 m/s at its maximum
height of *y* = 336 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?

Answer #1

a) initial total mechanical energy is KE + PE = (0.5*m*v^2)+(m*g*h) = (0.5*44*144^2)+(44*9.8*120)

Ei = 507936 J

b) vertival distance travelles is 336-120 = 216 m

at maximum height ,total mechanical energy is KE+PE = (0.5*44*102.1^2)+(44*9.8*336) = 374220.21 J

then required work done on projectile is 507936 - 374220.21 = 133715.79 J

C) Now work done when going down is 1.5*133715.79 = 200573.685 J

Total mechanical energy at the bottom is

374220.21 - (0.5*44*v^2)+(44*9.81*336) = 200573.685

v = 120.355 m/sec

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