Answer
$h=0.29m$
Work Step by Step
We know that
$K_i+U_i=K_f+U_f$
$\implies 0+mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2$
$mgh=\frac{1}{2}mv^2+\frac{1}{2}I(\frac{v}{r})^2$
$mgh=\frac{1}{2}v^2(m+\frac{1}{r^2})$
This can be rearranged as:
$h=\frac{v^2}{2mg}(m+\frac{1}{r^2})$
We plug in the known values to obtain:
$h=\frac{(0.65)^2}{2(0.056)(9.81)}(0.056+\frac{2.9\times 10^{-5}}{(0.0064)^2})$
$h=0.29m$
