# How high does a slide need to be if the person at the bottom grabs a pole and rotates the pole 72 degrees?

My attempt:

$mgh = \frac{1}{2}I \omega^2$

$\omega^2 = 2 \alpha\Delta \theta$, so

$mgh = \frac{1}{2}I 2 \alpha \Delta \theta$

$mgh = I \alpha \Delta \theta = \tau \Delta\theta$

$\tau = rF\sin \theta = Lma$ ( $L$ for the length of the string, $m$, the mass of the person, and $a$ for the person's linear acceleration)

$mgh = Lma\Delta\theta$

$h = \frac{Lma \Delta\theta}{mg} = \frac{La \Delta\theta}{g}$

I don't know what to do now since I'm not sure I can solve for $a$.