The ball is thrown from a certain height vertically down onto a horizontal platform at a speed of 20 m / s.

The ball is thrown from a certain height vertically down onto a horizontal platform at a speed of 20 m / s. How much above the initial level will the ball bounce? the impact is considered absolutely elastic

V0 = 20 m / s.

g = 10 m / s2.

Δh -?

Δh = h – h0, where h is the maximum height of the ball after bouncing, h0 is the height from which the ball is thrown.

If we consider the impact of the ball on the platform as absolutely elastic, then this means that the law of conservation of total mechanical energy is valid during the impact.

At the moment of the throw, the total mechanical energy of the ball consists of potential En0 and kinetic energy Ek0. When bouncing off the court at the maximum height, the ball stops, its total mechanical energy will consist only of the potential energy En.

En0 + Ek0 = En.

En0 = m * g * h0, Ek0 = m * V0 ^ 2/2, En = m * g * h.

m * g * h0 + m * V0 ^ 2/2 = m * g * h.

m * g * h – m * g * h0 = m * V0 ^ 2/2.

m * g * (h – h0) = m * V0 ^ 2/2.

h – h0 = m * V0 ^ 2/2 * m * g = V0 ^ 2/2 * g.

Δh = (20 m / s) ^ 2/2 * 10 m / s2 = 20 m.

Answer: the ball will bounce Δh = 20 m higher than the height of the throw.