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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 :

xero099

Answer:

[tex]F_{net} = 24434.783\,N[/tex]

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:

[tex]m_{person}\cdot v + Imp = 0[/tex]

The impact is:

[tex]Imp = - m_{person}\cdot v[/tex]

[tex]Imp = - (70\,kg)\cdot (-7.70\,\frac{m}{s} )[/tex]

[tex]Imp = 539\,N\cdot s[/tex]

The force of impact is derived as follows:

[tex]F_{net} = \frac{Imp}{\Delta t}[/tex]

[tex]F_{net} = \frac{539\,N\cdot s}{0.023\,s}[/tex]

[tex]F_{net} = 24434.783\,N[/tex]

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