If you stand on the top of a falling ladder you will hit the ground at a higher speed (and therefore presumedly sustain more injury) if you hold on to the ladder than if you jump off it. This was solved here.
Where is the "break even" height on the ladder, from where you will hit the ground with the same speed if you jump off it or if you follow it down? This question just makes an assumption that you would hit the ground more softly if you stay on to the ladder (compared to jumping off it), if you are located at the lower part of it.
I don't think the midpoint is the break even point. I quickly calculated and I think you should stay on the ladder if you are on its midpoint. (The following is just a quick computation, there could be errors in it.)
$$ v_{midpoint}^2 = \frac{1}{2}gl \frac{m + m_L}{\frac{1}{4}m + \frac{1}{3}m_L} $$
$m$ and $m_L$ are the respective masses of man and ladder, $l$ is the length of the ladder.
Answer
You should stand at 2/3 of the height of the ladder.
If you land with the same kinetic energy as without a ladder, then the ladder should land with the same kinetic energy as without you. Equating the kinetic energy of the ladder with its potential energy at the beginning:
$$\frac{1}{2} mgL = \frac{1}{2} I_L \omega^2 = \frac{1}{2} \left(\frac{1}{3} mL^2\right) \omega^2$$ gives: $$\omega = \sqrt{\frac{3g}{L}}$$
where $L$ is the length, $m$ the mass, $I_L$ the moment of inertia and $\omega$ is the angular velocity of the ladder.
For you the same equation holds, but now $\omega$ is known:
$$MgH = \frac{1}{2} I_M \omega^2 = \frac{1}{2} (MH^2) \left(\frac{3g}{L}\right)$$ with $M$ your mass, $I_M$ your moment of inertia and $H$ your height. Solving for $H$ gives:
$$H=\frac{2}{3}L$$
or of course $H=0\;.$
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