physics
posted by Austin on .
A person jumps from the roof of a house 4.0 high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.74. 1)If the mass of his torso (excluding legs) is 40 , find his velocity just before his feet strike the ground. 2)If the mass of his torso (excluding legs) is 40 , find the magnitude of the average force exerted on his torso by his legs during deceleration.

mgh=1/2 m v^2
v=sqrt 4*g
2) force*distance=1/2 m v^2
force= mass*(4g)/2d= 2*g*mass/.74
check that. 
I don't understand this part....
2) force*distance=1/2 m v^2
force= mass*(4g)/2d= 2*g*mass/.74