Question

A 70.0 kg person jumping down from a 3.00 meter wall hits the ground at 7.70 m/s. He hits the ground with stiff legs and stops in 0.023 seconds. What is the force of impact, in newtons?

Answer 1

Answer:

$F_{net} = 24434.783\,N$

Explanation:

Let consider that the person hit the ground in the negative direction. The physical model for the impact is modelled after the Impact Theorem:

$m_{person}\cdot v + Imp = 0$

The impact is:

$Imp = - m_{person}\cdot v$

$Imp = - (70\,kg)\cdot (-7.70\,\frac{m}{s} )$

$Imp = 539\,N\cdot s$

The force of impact is derived as follows:

$F_{net} = \frac{Imp}{\Delta t}$

$F_{net} = \frac{539\,N\cdot s}{0.023\,s}$

$F_{net} = 24434.783\,N$