Question 58: Motion in 1D

Question

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 + 2e
2
h + 2e
4
h + 2e
6
h + 2e
8
h + ......
d = h + 2e
2
h (1 + e
2
+ e
4
+ e
6
+ .......)
$$d = {{(1 - {e^2})h + 2{e^2}h} \over {1 - {e^2}}} = {{h(1 + {e^2})} \over {1 - {e^2}}}$$
Total time = T + 2eT + 2e
2
T + 2e
3
T + .......
Total time = T + 2eT (1 + e + e
2
+ e
3
+ .......)
= $$T + 2e.T\left( {{1 \over {1 - e}}} \right)$$
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.

Answer

2.50 ms$$-$$1

Answer

3.0 ms$$-$$1

Answer

2.0 ms$$-$$1

Answer

3.50 ms$$-$$1

Motion in 1D

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)

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