A 50.0-kg projectile is fired at an angle of 30.0 degrees above the horizontal w

Question

A 50.0-kg projectile is fired at an angle of 30.0 degrees above the horizontal with an initial speed of 1.20 x 10^2 m/s from the top of a cliff 142 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 85.0 m/s at its maximum height of y = 427 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) initial total mechanical energy = mgh+.5*m*v^2 =429.6J
b)
work done on the projectile by air friction= 429.6 kJ-(50*9.8*427+.5*50*85^2) = 39.8kJ
c)speed =sqrt(2*(429.6-59.7)*10^3 /50) =121.6 m/s

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