A ball of mass 20 g falls from a height of 45 m. It rebounds from the ground to a height of 40 m. Calculate:
(a) initial potential energy of the ball.
(b) the speed of the ball at which it hits the ground.
(c) the loss in kinetic energy on striking the ground.
[g = 10 m s−2]
Given, Mass (m) = 20 g = 0.02 kg,
Initial height (h1) = 45 m,
Rebound height (h2) = 40 m and g = 10 m s−2.
(a) Initial potential energy = mgh1 = 0.02 × 10 × 45 = 9 J.
(b) Just before hitting the ground:
\frac{1}{2}\mathrm{mv^2}=9v = \sqrt{900} = 30 m s−1.
(c) Kinetic energy after rebound = 0.02 × 10 × 40 = 8 J.
Loss in kinetic energy = 9 − 8 = 1 J.
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