QJEEM21Ph9a9e: JEE Main 2021 Physics - Class 11 JEEM21Ph9a9e Solution: A rubber ball is released from a height of 5 m above the floor. …

Question

Verified Solution • 2025–26 Session moderate

A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to

{{81} \over {100}}

of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms^{$-$ 2})

Accepted Answer

Total distance d = h + 2e²h + 2e⁴h + 2e⁶h + 2e⁸h + ......

d = h + 2e²h (1 + e² + e⁴ + e⁶ + .......)

d = {{(1 - {e^2})h + 2{e^2}h} \over {1 - {e^2}}} = {{h(1 + {e^2})} \over {1 - {e^2}}}

Total time = T + 2eT + 2e²T + 2e³T + .......

Total time = T + 2eT (1 + e + e² + e³ + .......)

=

T + 2e.T\left( {{1 \over {1 - e}}} \right)

Total time

= {{T(1 + e)} \over {1 - e}}

Average speed of the ball

{V_{avg}} = {{h{{(1 + {e^2})} \over {(1 - {e^2})}}} \over {T\left( {{{1 + e} \over {1 - e}}} \right)}}

= {5 \over 1}\left( {{{1 + {e^2}} \over {(1 + e)(1 - e)}}{{(1 - e)} \over {(1 + e)}}} \right)

{V_{avg}} = {{5(1 + {e^2})} \over {{{(1 + e)}^2}}}

\because

{h^1} = {e^2}h

{{81} \over {100}} = {e^2}

e = {9 \over {10}} = 0.9

{V_{avg}} = {{5\left( {1 + {{81} \over {100}}} \right)} \over {{{(1 + 0.9)}^2}}}

= 2.50 m/sec.

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