Asked by Austin
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.
Answers
Answered by
bobpursley
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.
v=sqrt 4*g
2) force*distance=1/2 m v^2
force= mass*(4g)/2d= 2*g*mass/.74
check that.
Answered by
Emily
I don't understand this part....
2) force*distance=1/2 m v^2
force= mass*(4g)/2d= 2*g*mass/.74
2) force*distance=1/2 m v^2
force= mass*(4g)/2d= 2*g*mass/.74
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