When should you jump off a falling ladder?

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 $H=0\;.$

We can do this with a minor modification to the calculation described in the earlier question. As before we'll take the ladder length to be $\ell$, but now we'll take your height to be $\alpha\ell$, where $\alpha$ ranges from zero to one. Our reference point is if you let go, in which case your speed when you hit the ground will be:

$$ v^2 = 2g\alpha\ell \tag{1} $$

Now suppose you hold onto the ladder. As before we calculate the total potential energy change of both you and the ladder, which is:

$$ V = mg\alpha\ell + \frac{1}{2}m_Lg\ell \tag{2} $$

And this must be equal to the increase in angular kinetic energy $\tfrac{1}{2}I\omega^2$. The combined moment of inertia of you and ladder is:

$$ I = m(\alpha\ell)^2 + \frac{1}{3}m_L\ell^2 $$

And setting the kinetic energy equal to the potential energy gives:

$$ mg\alpha\ell + \frac{1}{2}m_Lg\ell = \tfrac{1}{2}\left(m\alpha^2 + \frac{m_L}{3}\right)\ell^2\omega^2 $$

And since $v=r\omega$ your velocity is $v=\alpha\ell\omega$ giving:

$$ mg\alpha\ell + \frac{1}{2}m_Lg\ell = \tfrac{1}{2}\left(m\alpha^2 + \frac{m_L}{3}\right)\ell^2\frac{v^2}{\alpha^2\ell^2} $$

Which rearranges to:

$$ v^2 = g\ell\alpha^2 \frac{2m\alpha + m_L}{m\alpha^2 + \frac{m_L}{3}} $$

And finally substitute for $v$ from equation (1) to get:

$$ 2g\alpha\ell = g\ell\alpha^2 \frac{2m\alpha + m_L}{m\alpha^2 + \frac{m_L}{3}} $$

And this rearranges to:

$$ \alpha = \frac{2}{3} $$

So if you are more than $\tfrac{2}{3}$ of the way up the ladder you should let go, while if you are lower than $\tfrac{2}{3}$ of the way up the ladder you should hang on.