A hollow basketball, with radius 13.1 cm and mass 0.260 kg, is released from rest at the top of a 3.60 m long ramp inclined at 26.0° above the horizontal. What is the ball's velocity when it reaches the bottom of the ramp?

Answer 1

figure out how far it fell
like
h = 3.6 sin 26.0
then the potential energy at the top = m g h
That will be the kinetic energy at the bottom
m g h = (1/2)m v^2 + (1/2) I omega^2
but v = omega*R
so
m g h = (1/2) m v^2 + (1/2) I (v^2/R^2)

I is the moment of inertial of a thin walled sphere of radius 0.131 and mass 0.260 about an axis through the center.

Answer 2

ah, note, you do not really need to know the mass :)

Answer 3

because I =(2/5) m R^2

m g h = (1/2) m v^2 + (1/2) (2/5)m R^2 (v^2/R^2)
g h = (1/2) v^2 + (1/5) v^2 = 0.7 v^2
v^2 =( g* 3.6 sin 26 )/ 0.7