Velocity of hitting the ground when falling forward A friend of mine fell while he was snowboarding which led to a discussion. He mentioned that, in general, if your center of mass is traveling at a certain velocity, and you fall forward, your head would contact the ground at twice the velocity that your center of mass was traveling. Is this true? And if so can you explain the physics behind it to me?
 A: If you assume that


*

*Your body is a uniform, thin rigid rod.

*One end of the rod is pivoted (aka your feet) during the fall.
Then one simply recalls that the angular velocity $\omega$ of rotation of your body is related to the tangential velocity $v$ of a point a distance $r$ from the pivot by
\begin{align}
  v = \omega r
\end{align}
Now, if you have height $h$, then your center of mass will be a distance $h/2$ from your feet, and your head will be a distance $h$ from your feet, so the ratio $v_\mathrm{head}/v_\mathrm{COM}$ of the tangential velocity of your head to that of your center of mass is
\begin{align}
  \frac{v_{\mathrm{head}}}{v_{\mathrm{COM}}} = \frac{\omega h}{\omega (h/2)} = 2.
\end{align}
In other words, the tangential velocity of your head will be twice as large as that of your center of mass.
The intuition behind this is simply that for a given change in the angular position of your body, your head moves twice as far as your center of mass.
Note that in the real world, the assumptions above aren't really correct, but they do give you an idea of the basic physics in a rough approximation.
