A ball of mass 0.2 kg is dropped from a height of 20 m . Upon...

Velocity of the ball just before striking the ground:
$v_1 = \sqrt{2gh_1} = \sqrt{2 \times 10 \times 20} = 20 \text{ m s}^{-1}$ (downwards)

Taking the upward direction as positive, the initial momentum of the ball is:
$p_i = m(-v_1) = 0.2 \times (-20) = -4 \text{ kg m s}^{-1}$

According to the impulse-momentum theorem:
$J = p_f - p_i$
$12 = p_f - (-4)$
$12 = p_f + 4 \Rightarrow p_f = 8 \text{ kg m s}^{-1}$

The rebound velocity $v_2$ is:
$v_2 = \dfrac{p_f}{m} = \dfrac{8}{0.2} = 40 \text{ m s}^{-1}$ (upwards)

The height to which the ball rebounds is:
$h_2 = \dfrac{v_2^2}{2g} = \dfrac{40^2}{2 \times 10} = \dfrac{1600}{20} = 80 \text{ m}$

Answer: $80 \text{ m}$