I\'m standing on top of a 10 meter high building holding a 7 kg bowling ball. (a

Question

I'm standing on top of a 10 meter high building holding a 7 kg bowling ball. (a) What's the potential energy of the bowling ball? (b) What's its kinetic energy? I see my professor walking by and decide to drop the ball on his head. Assume the professor is 1.8 meters tall. (c) What is the ball's kinetic energy when it hits his head? (d) What's its potential energy at this point? (e) How fast is the ball traveling when it hits the professor's head? (f) With what force does the ball hit the professor's head? (g) What's the velocity of the ball halfway to the professor's head? (h) If the professor looks up just as you release the bowling ball, how long does he have to move out of the way before the ball hits his head?

Explanation / Answer

(a)the potential energy U = m * g * h m = 7 kg,g = 9.8 m/s^2 and h = 10 m (b)the kinetic energy K = (1/2)m * v^2 v = 0 or K = (1/2)m * (0)^2 = 0 (c)kinetic energy K1 = (1/2)m * v^2 v^2 = 2 * g * h1 h1= (10 - 1.8) m = 8.2 m h1= (10 - 1.8) m = 8.2 m (d)potential energy U1= m * g * h1 (e)speed v = (2 * g * h1)^(1/2) (f)force F = m * g (h)time is v = u + gt u = 0 or t = (v/g) = ((2 * g * h1)^(1/2)/g) = (2 *h1/g)^(1/2)

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