## Essential University Physics: Volume 1 (3rd Edition)

We use conservation of energy to find: $mgh = \frac{1}{2}mv^2+I\omega^2$ $mgh = \frac{1}{2}mv^2+I\frac{v^2}{r^2}$ $h = \frac{\frac{1}{2}mv^2+I\frac{v^2}{r^2}}{mg}$ $h = \frac{\frac{1}{2}(395)(23.6)^2+(2.1)\frac{23.6^2}{.26^2}}{(395)(9.81)}=\fbox{32.85 meters}$