A ball of mass 0.5kg is dropped from a height of 10m and rebound with a velocity 1/3 of that before impact. The height reached after rebound is?

Answer 1

It's KEnergy is now (1/3)^2 the original (KE depends on velocity squared), so it can only go to1/3 the original height.

Answer 2

V^2 = Vo^2 + 2g*h = 0 + 19.6*10 = 196,
V = 14 m/s before impact.

Vr = V/3 = 14/3 = 4.67 m/s = Velocity at which the ball rebounds.

V^2 = Vr^2 + 2g*h = 0.
4.67^2 + (-19.6)h = 0,
h =