# Calculating the force of someone falling backwards

Say someone was standing and fell backwards. Knowing their mass and height, would it be possible to calculating the force they hit the ground with, not accounting for air resistance?

What I've tried:

Say $h$ is the height of the person, in meters and $m$ is their mass. The distance of their fall would be $\frac{2h*\pi}{4}$. Let's call that $d$.

I know that $V=\sqrt{2dg}$. So, with $h=1.68%$

$d$ then equals $2.63$ And $g$ on earth is $9.81 m/s^2$, so, plugging in variables we get: $V=\sqrt{2\cdot 2.63\cdot 9.81} = 7.195$

I have 3 questions. First, is what I have done so far correct? Second, what are the units that the velocity is in? And finally, what do I do from here to get the force?

• Calculating impact forces is tricky because they depend on the properties of the ground; imagine someone falling onto concrete vs falling onto a mattress. Dec 9, 2015 at 22:58
• (and the properties of the falling bodies). Dec 9, 2015 at 23:25
• @Javier So is it possible to even approximate using only mass and height? Dec 9, 2015 at 23:29

Where you invoked:

$$V=\sqrt{2dg},$$

you were referring to the conversion of potential energy to kinetic energy acc.:

$$\frac{1}{2}mV^2=mgd$$

But $\frac{2h\pi}{4} \neq d$. In terms of potential energy it is simply the displacement in the sense of the gravitational field and not the length travelled, that is:

$$d=\frac{h}{2},$$

$$V=\sqrt{hg}.$$
Because $g$ is expressed in $\mathrm{m/s^2}$ and $h$ in $\mathrm{m}$, then $hg$ is in $\mathrm{g^2/s^2}$. Take the square root of that and you get $\mathrm{m/s}$, the standard unit for velocity.
$$F=ma.$$
Imagine falling on a pile of soft cushions: you'll decelerate more gently (smaller $a$) than when falling on a hard concrete floor (larger $a$). The force $F$ experienced in the latter case will be higher than in the former case.