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.

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