To what height will a vertically falling ball weighing 0.4 kg bounce off after an absolutely elastic

To what height will a vertically falling ball weighing 0.4 kg bounce off after an absolutely elastic impact on a horizontal plane, if the change in momentum is 8 kg * m / s?

m = 0.4 kg.

Δp = 8 kg * m / s.

g = 10 m / s.

h -?

Before the ball hit the horizontal plane, its velocity was directed vertically downward. After an absolutely elastic impact, its velocity will be directed vertically upward. Let us express the change in impulse Δp in vector form: Δp = m * V2 – m * V1, where V2, V1 are the velocity of the ball after and before the impact.

With an absolutely elastic impact, the velocities are equal in modulus V1 = – V2 and oppositely directed.

For projections onto the vertical axis: Δp = m * V2 – m * (- V1) = 2 * m * V2.

V2 = Δp / 2 * m.

According to the law of conservation of total mechanical energy: m * g * h = m * V2 ^ 2/2.

h = m * V2 ^ 2/2 * m * g = V2 ^ 2/2 * g = Δp2 / 8 * m2 * g.

h = (8 kg * m / s) ^ 2/8 * (0.4 kg) ^ 2 * 10 m / s = 1.25 m.

Answer: the ball will bounce to a height of h = 1.25 m.