Answer:
K = (1/2) mv² + (1/2) Iω², where m is the ball mass, I is the ball's moment of inertia (2/5)mr², and ω is the angular velocity of the ball. Because the ball rolls without slipping, it is easy to see that v=ωr, or r=v/ω. Then,
K = (1/2)mv² + (1/2)(2/5)mr²ω²
= (1/2)mv² + (1/5)mv²
= (7/10)mv²
Setting potential at the top equal to kinetic at the bottom,
mgh=(7/10)mv²
v=âš{(10/7)(gh)}
= [(10/7)(9.8)(0.51)]^(1/2) = 2.672m/s
