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If someone falls from a sixth storey building, with what force will he land? By this I mean to say that if he weighs 50 kg and was initially at rest before falling, which implies it was a free fall. We can easily calculate the velocity and momentum as we know the acceleration due to gravity... but with what force he will land?

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  • $\begingroup$ f = ma, calculate the velocity then assume he decelerates to zero in a tenth of a second for a rough estimate $\endgroup$ – Alex Robinson Apr 8 '17 at 19:55
  • $\begingroup$ Depends how soft and bouncy he is :) $\endgroup$ – milo Apr 8 '17 at 20:09
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It is the not the $mv$ that kills you, but the $\frac{d p}{dt}$.

The ground will eventually bring you to rest, in a time $\delta t$. This can be very large for soft materials, like a bouncy castle or sponge, or very short for incompressible materials like concrete and steel.

Your momentum change $\Delta p$ from the instant before impact and rest is $0 - mv_{final} = - mv_{final}$. Hence, the ground exerts a force $F$ on you: $$ F = \frac{dp}{dt} \approx -\frac{\Delta p}{\delta t},$$ which is negative as it needs to decelerate you to bring you to rest.

$v_{final}$ is calculated with the usual suvat equations.

What is the actual physical nature of the force? The resistance of objects to being deformed, hence the $\delta t$ dependence that I mentioned earlier. In essence this boils down to molecular bonds so it's electromagnetism.

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