Ip a solid sphere with a diameter of 0.19 m is released from rest; it then rolls without slipping down a ramp, dropping through a vertical height of 0.61 m. the ball leaves the bottom of the ramp, which is 1.22 m above the floor, moving horizontally.

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