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 $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )
Correct Option: , 4
(4) $v_{0}=\sqrt{2 g h}$
$\mathrm{v}=\mathrm{e} \sqrt{2 \mathrm{gh}}=\sqrt{2 \mathrm{gh}}$
$\Rightarrow \mathrm{e}=0.9$
$S=h+2 e^{2} h+2 e^{4} h+\ldots \ldots \ldots$
$\mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}+2 \mathrm{e} \sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}+2 \mathrm{e}^{2} \sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}+\ldots \ldots \ldots$
$\mathrm{v}_{\mathrm{av}}=\frac{\mathrm{s}}{\mathrm{t}}=2.5 \mathrm{~m} / \mathrm{s}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.