I asked this question earlier, but no one answered... please help.?

Question

A 67.6 kg man steps off a platform 3.68 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 0.39 m before coming to rest. Treating our rigid legged friend as a particle, what is the average force his feet exert on the ground while he slows down?

Note: Assume the acceleration while he is slowing down is constant.

Give your answer in Newtons to the nearest whole number.

Hint: Draw a free body diagram to aid in seeing the forces.

Answer 1

The man starts from rest so initial velocity u = 0. The man has a velocity v1 when he just touches the ground.
When he touches the ground, he retards for a further distance of 0.39m before coming to rest. So the second part (retardation) the initial velocity is v1 and final velocity is 0.

First part.

V^2 = u^2 + 2 * 9.8*3.68
V^2 = 72.128
so, v1 = 8.49 m/s

Second part
0 = 8.49^2 - 2*a*0.39
0= 72.128-2*a*0.39
a = 92.47 m/ s^2


Force = m*a
=67*6*92.47 N
6250.97 N

Another way of answering -

The distance travelled = 3.68 - 0.39 = 3.29 in part 1. and proceed as above.
(This is because, we do not know whether .39 m is included in the distance to fall or not.)

This time you have the answer.

Answer 2

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Answer 3

Probably no one answered because they respect you enough to let you do your own homework.