Question:
A rubber ball is released from a height of $5 \mathrm{~m}$ 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 $g=10 \mathrm{~ms}^{-2}$ )
Correct Option: , 4
Solution:
(4)
$v_{0}=\sqrt{2 g h}$
$v=e \sqrt{2 g h}=\sqrt{2 g h}$
$\Rightarrow e=0.9$
$s=h+2 e^{2} h+2 e^{4} h+\ldots \ldots \ldots$
$t=\sqrt{\frac{2 h}{g}}+2 e \sqrt{\frac{2 h}{g}}+2 e^{2} \sqrt{\frac{2 h}{g}}+\ldots \ldots \ldots . .$
$v_{a v}=\frac{s}{t}=2.5 \mathrm{~m} / \mathrm{s}$
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