Starting from rest, a basketball rolls from the top of a hill to the bottom, reaching a translational speed of 6.12 m/s. Ignore frictional losses.
(a) What is the height of the hill?
(b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?
The Attempt at a Solution
Initial Total mechanical energy = Final Total mechanical energy
1/2mv^2 + 1/2IW^2 + mgh = 1/2mv^2 + 1/2IW^2 + mgh
Initial KE = 0 and final PE = 0
mgh = 1/2mv^2 + 1/2IW^2
gh = 1/2v^2 + 1/2IW^2
I am stuck with 2 unknown variables W and h.