- #1

HJ^2

- 44

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## Homework Statement

A 74.4 kg person falls straight down from a 1.96 m height (measured from his feet) and lands with weight distributed equally on both feet. To soften the blow, he bends his knees so that it takes 0.906 s for him to stop once his feet touch the ground. What constant force does the floor exert on each foot while he's stopping? Give your answer in Newtons

## Homework Equations

f=ma

f=(m*v^2) / 2x

F * t = m * Δv

Δp=m(v-vo)

Favg=Δp/Δt

w=mg

Xf=Xi+Vit+(at^2/2)

Vf=Vi+at

Vf^2=Vi^2+2a(Xf-Xi)

## The Attempt at a Solution

The answer is 254.5 and I am trying to work backwards to see what I am doing incorrectly.

Originally, I had lumped all variables into one equation ( f * t = m * Δv ) but now I think it's two parts? First system being A; person falling 1.96m and system B; ground receiving force and pushing back up as the person absorbs shock @ 0.0906s?

I

*think*the equation might be solved by calculating acceleration and solving for force with Newtons 2nd law, but I am unsure and that sounds completely wrong.

Thanks in advance