A ball is allowed to fall from a height of 15 m. Calculate the velocity with which it strikes the ground. To what height will be the ball re bounced if it loses 40% of its energy on striking the ground.

Answer 1

v = √2as = √(2*9.8*15) = 17.146 m/s

Since the ball's energy

PE = mgh, if it loses 40% of its energy, it will lose 40% of its height, so it will bounce to 9 m.

Or, since PE becomes KE just before impact,

KE = 1/2 mv^2
The rebound velocity will be

.6 * 1/2 m*17.146^2 = 1/2 mv^2
v = √.6 * 17.146 = 13.281 m/s

So, the max height is

h = v^2/2g = 8.999 ≈ 9 m