Starting from rest, a basketball rolls from the top of a hill to the bottom, reaching a translational speed of 3.83 m/s. Ignore frictional losses.

Question

(a) What is the height of the hill?

I've tried to use the mgh=1/2mv^2. I think it's because it's translational speed. I just need the proper formula that's all. I'm entirely sure where to start.

Answer 1

it has not only (1/2) m v^2
but also (1/2) I w^2
where w = angular velocity
if it does not slip then w = v/r

I = (2/3) m r^2 for thin spherical shell

so
ke = (1/2) m v^2 + (1/2) m (2/3) r^2 v^2/r^2

mgh = (1/2) m ( v^2 + 2/3 v^2)
h = (1/2g) (5/3) (3.83)^2