Part a) What is his speed at the instant his feet touch the ground? v= ? m/s Par

Question

Part a)

What is his speed at the instant his feet touch the ground? v= ? m/s

Part b)

A 72.8 kg man steps off a platform 3.07 m above the ground. He keeps his legs straight as he falls, but at the moment his feet touch the ground his knees begin to bend, and, treated as a particle, he moves an additional 0.620 m before coming to rest. Part a) What is his speed at the instant his feet touch the ground? v= ? m/s Part b) Treating him as a particle, what is the magnitude of his acceleration as he slows down if the acceleration is constant? a= ? m/s^2 Part C) Treating him as a particle, what is the direction of his acceleration as he slows down if the acceleration is constant? upward or downward?

Explanation / Answer

a)the speed at the time his feet touch the ground
v = (2gh)
= (2 * 9.81 * 3.07)
= 7.76 m/s

b)As he slows down, lets assume his deceleration is a
v2 - u2 = 2as
a = - u2 / 2s
a = - 7.76 * 7.76 / 2 * 0.620
a = - 48.56 m/s2
c) It sign is negative, because it is opposing of the motion.direction of acceleration upwards

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