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?
 A: 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:
$$d=\frac{h}{2},$$
so that in your case:
$$V=\sqrt{hg}.$$

Second, what are the units that the velocity is in? And finally, what do I do from here to get the force?

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.
Re. you last question, that depends what you're falling onto. The force you'll be experiencing is due to the deceleration the object undergoes acc.:
$$F=ma.$$
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.
