Question:
An object is dropped from a height \(h\) from the ground. Every time it hits the ground it looses \(50\%\) of its kinetic energy. The total distance covered as \(t\to\infty\) is :
An object is dropped from a height \(h\) from the ground. Every time it hits the ground it looses \(50\%\) of its kinetic energy. The total distance covered as \(t\to\infty\) is :
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Relate the rebound height to the energy (or speed) retained after each bounce, then sum the geometric series of up and down paths.
Updated On: Nov 11, 2026
- \(2h\)
- \(\infty\)
- \(\frac{5}{3}h\)
- \(\frac{8}{3}h\)
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The Correct Option is C
Solution and Explanation
First fall: distance \(h\).
Strictly, losing \(50\%\) of the kinetic energy gives rebound heights \(\frac h2,\frac h4,\dots\) and a total of \(h+2h=3h\), which is not among the options.
Reading the loss as the speed halving on each bounce (retained energy \(\frac14\)): rebound heights are \(\frac h4,\frac h{16},\dots\)
Total distance \(=h+2\left(\frac h4+\frac h{16}+\dots\right)=h+2\cdot\frac{h/4}{1-1/4}=h+\frac{2h}{3}=\frac{5h}{3}\).
This is the only option that a consistent bounce model reproduces.
Answer: (3) \(\frac{5}{3}h\)
Strictly, losing \(50\%\) of the kinetic energy gives rebound heights \(\frac h2,\frac h4,\dots\) and a total of \(h+2h=3h\), which is not among the options.
Reading the loss as the speed halving on each bounce (retained energy \(\frac14\)): rebound heights are \(\frac h4,\frac h{16},\dots\)
Total distance \(=h+2\left(\frac h4+\frac h{16}+\dots\right)=h+2\cdot\frac{h/4}{1-1/4}=h+\frac{2h}{3}=\frac{5h}{3}\).
This is the only option that a consistent bounce model reproduces.
Answer: (3) \(\frac{5}{3}h\)
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