How does vertical deformation of an object soften the free fall of a body?

Question

I'm a bit confused about the situation in this exercise:

A man fell off a building of height $h$, with null initial velocity, but he survived thanks to a metal box that softened the fall deforming vertically by 50 cm. Calculate the acceleration (assumed constant) experienced by the man, in terms of $g$.

Now, I guess this is not a trick question (i.e. the acceleration is $g$) as it sort of makes sense: reaching the ground directly would have been worse for the man, because it does not deform. Also, a trampoline is in turn better than a metal box because it's more elastic. So it seems that the acceleration at the moment of arrival differs. But how does it depend on the vertical deformation $d$?

Answer 1

Forces applied to a human being can do damage to the body.
The larger the forces the more damage is done.

If the forces can be reduced then less damage is done.

Suppose the speed of the body, mass $m$, just before hitting an obstacle is $v$ and after hitting the obstacle the body is at rest.

The magnitude of the change of momentum of the body is $mv$.

To change the momentum of the body an average force $F$ must be applied to the body over a time $t$. Using Newton’s second law the force which must be applied to the body is $F = \dfrac {mv}{t}$.

This expression for the applied force tells you that for a given change in momentum the longer the time taken for the body to slow down the smaller is the force applied on the body which is equivalent to a softer landing.

In your example instead of stopping in a very small distance when hitting a concrete floor which takes a very short period of time hitting the box means that the slowing down time over a distance of $50\, \rm cm$ is larger and the force on the body correspondingly smaller (equivalent to a softer landing).

You can use constant acceleration kinematic equations to find the acceleration during the time the can crumples by $50\,\rm cm$ after the body has fallen from rest a distance $h \,\rm cm$.

Answer 2

When landing the momentum is reduced from $m\sqrt{2sg}$ to zero over a time interval $\Delta t$. The force on the man will be on average $m\sqrt{2sg} /\Delta t$, which is smaller if $\Delta t$ is larger. A metal box that is flattened or a trampoline prolong the duration of the impact and therefore reduce the force, increasing the chance of survival.

Cars have a so called "crumple zone" for the same reason: to limit the g forces in case a crash occurs.