Question:
A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to \(\frac{81}{100}\) of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms-2)
A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to \(\frac{81}{100}\) of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms-2)
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Height after each bounce falls by \(e^2 = 0.81\) so \(e = 0.9\); sum the infinite geometric series of distances and of times, then divide.
Updated On: Nov 4, 2026
- \(2.50\ \mathrm{ms^{-1}}\)
- \(3.0\ \mathrm{ms^{-1}}\)
- \(2.0\ \mathrm{ms^{-1}}\)
- \(3.50\ \mathrm{ms^{-1}}\)
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The Correct Option is A
Solution and Explanation
Height ratio \(e^2 = \frac{81}{100}\), so the speed ratio after/before each bounce is \(e = 0.9\).
Distance: \(D = 5 + 2(5)(0.81) + 2(5)(0.81)^2 + \dots = 5 + \frac{2(5)(0.81)}{1 - 0.81} = 5 + \frac{8.1}{0.19} = 47.63\ \mathrm{m}\).
Time: first fall \(t_0 = \sqrt{\frac{2\times5}{10}} = 1\ \mathrm{s}\). Each later up-and-down takes \(2t_0e^n\): \(T = 1 + 2(0.9) + 2(0.9)^2 + \dots = 1 + \frac{1.8}{1 - 0.9} = 19\ \mathrm{s}\).
Average speed \(= \frac{D}{T} = \frac{47.63}{19} = 2.507 \approx 2.50\ \mathrm{m\,s^{-1}}\).
Answer: (1) \(2.50\ \mathrm{ms^{-1}}\)
Distance: \(D = 5 + 2(5)(0.81) + 2(5)(0.81)^2 + \dots = 5 + \frac{2(5)(0.81)}{1 - 0.81} = 5 + \frac{8.1}{0.19} = 47.63\ \mathrm{m}\).
Time: first fall \(t_0 = \sqrt{\frac{2\times5}{10}} = 1\ \mathrm{s}\). Each later up-and-down takes \(2t_0e^n\): \(T = 1 + 2(0.9) + 2(0.9)^2 + \dots = 1 + \frac{1.8}{1 - 0.9} = 19\ \mathrm{s}\).
Average speed \(= \frac{D}{T} = \frac{47.63}{19} = 2.507 \approx 2.50\ \mathrm{m\,s^{-1}}\).
Answer: (1) \(2.50\ \mathrm{ms^{-1}}\)
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