Question

A ball falling freely from a height of 4.9 m/s, hits a horizontal surface. If $e = \frac{3}{4}$, then the ball will hit the surface, second time after.

• A. 1.0 s
• B. 1.5 s
• C. 2.0 s
• D. 3.0 s

Answer 1

Answer: (B)

1.5 s

Answer 2

The ball falls from 4.9 m, taking $t_1 = \sqrt{2 \times 4.9 / 9.8} = 1$ s to hit the ground. The velocity before the first hit is $v_1 = 9.8 \times 1 = 9.8$ m/s. After the bounce, the velocity is $v_1' = e v_1 = \frac{3}{4} \times 9.8$ m/s upwards. The time taken for the ball to go up and come down for the second flight (time between first and second hit) is $t_2 = \frac{2 v_1'}{g} = \frac{2 \times \frac{3}{4} \times 9.8}{9.8} = 2 \times \frac{3}{4} = 1.5$ s. Assuming the question asks for the time interval between the first and second hit, the answer is 1.5 s.