Somebody help me!!!?

Question

A person jumps from the roof of a house 3.9 m hight. When he strikes the ground below, he bends his knees so that his torso decelerates over an aproximate distance of 0.70 m. If the mas of his torso(excluding legs) is 42 kg, find a) his velocity just before his feet strike the ground, and b) the average force exerted on his torso by his legs during decceleration

Answer 1

First, the velocity just before the feet strike assuming the motion is directly downward:

Using conservation of energy:

m*g*3.9=.5*m*v^2
v=sqrt(2*g*3.9) m/s
=8.75 m/s

The work done by the force on the torso by the legs is:

W=F*d
=F*0.70

This is equal to the kinetic energy of the torso before the feet touch:

.5*42*2*g*3.9
F=42*g*3.9/0.70
=2300 N

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