Question

A person jumps from the roof of a house 2.6- m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 m . If the mass of his torso (excluding legs) is 41 kg.

Find his velocity just before his feet strike the ground.

Find the average force exerted on his torso by his legs during deceleration.

Answer 1

Height of roof ,H=2.6 m

Distance,d=0.70 m

mass of his torso,m=41 kg

Part a.

Velocity just before his feet strike the ground is

v=(2gH)^1/2=(29.82.6)^1/2=7.14 m/s

Part b.

Acceleration is given by

a=v²/2d=7.14²/(20.70)=36.4 m/s²

Average force=ma=4136.4=1492.4 N