Question

A person jumps from the roof of a house 3.5-m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.71 m .
If the mass of his torso (excluding legs) is 45 kg , find the magnitude of the average force exerted on his torso by his legs during deceleration.

Answer 1

The force exerted on his torso by his legs during the deceleration is 4365 N.

Explanation:

Mass of the torso m=45kg

Height of the building s=3.5 m

Decelerating distance=0.71 m

when he jumps to the ground, the only acceleration is acceleration due to gravity g

motion1 from top to ground

initial velocity u=0

we have to calculate final velocity v using the following equation of motion.

$v^2-u^2=2gs\\v^2-0^2=2\times 9.8\times3.5=68.6\\v=\sqrt{68.6} \\=8.3$

use height of the building as the distance s as the jump from top to the ground is only described here.

Motion 2 on the ground

v=0

u=8.3(final  velocity of motion 1)

The deceleration after striking the ground can  be calculated from the equation of motion

$v^2-u^2=2as\\\\a=v^2-u^2/2\times 0.71\\=0^2-8.3^2/0.71=97 m/s^2$

The decelerating distance is used in the place of s since since the motion after hitting the ground is described in this case.

The equation of force is

$F=ma\\=45\times 97=4365 N$