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

Question

A 48.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 1.32 102 m/s from the top of a cliff 148 m above level ground, where the ground is taken to be y = 0. (a) What is the initial total mechanical energy of the projectile? (b) Suppose the projectile is traveling 93.5 m/s at its maximum height of y = 329 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

(a) the initial total mechanical energy of the projectile
= 0.5 *48 * 132^2
= 418176 J --answer

(b) Suppose the projectile is traveling 93.5 m/s at its maximum height of y = 329 m.
height travelled = 329 - 148 = 181m
PE = 48* 9.81 * 181 = 85229.28 J
KE = 0.5 * 48 * 93.5^2 = 209814 J

work has been done on the projectile by air friction
= 418176 - 85229.28- 209814
= 123132.72 J ----answer

(c) 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
work done by friction = 1.5 * 123132.72 J = 184699.08 J

PE lost = 48 * 9.81 * 329 = 154919.52 J
KE gain = 0.5 * 48* v^2

418176 = 184699.08 + 154919.52 + 0.5 * 48 * v^2
v =57.21 m/s
answer

speed of the projectile immediately before it hits the ground

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