Question

A person jumps from the roof of a house 3.9-m

m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.75 m

m .

a)If the mass of his torso (excluding legs) is 45 kg

kg , find his velocity just before his feet strike the ground.

b)If the mass of his torso (excluding legs) is 45 kg

kg , find the magnitude of the average force exerted on his torso by his legs during deceleration.

c)What is the direction of this force? upward or downward

Answer 1

Explanations & Calculations

a)

$\qquad\qquad \begin{aligned} \small mgh&=\small \frac{1}{2}mv^2\\ \small v&=\small \sqrt{2gh} \end{aligned}$

b)

• The kinetic energy he has just before hitting the ground is expended against the force form the ground on him over the distance of deceleration. Therefore,

$\qquad\qquad \begin{aligned} \small \frac{1}{2}mv^2&=\small (F-mg)s\\ \small F&=\small mg+\frac{mv^2}{2s}\\ &=\small m(g+\frac{v^2}{2s}) \end{aligned}$

c)

• Since the force is from the ground on him, it is upwards.