A rubber ball is released from a height of $$5$$ m above the floor. It bounces back repeatedly, always rising to $$81100$$ of the height through which it falls. Find the average speed of the ball. (Take$$ $$g=10$$ ms−$$2$$ $$)

Question

A rubber ball is released from a height of $$5$$ m above the floor. It bounces back repeatedly, always rising to $$81100$$ of the height through which it falls. Find the average speed of the ball. (Take$$ $$g=10$$ ms−$$2$$ $$)

Infinity Learn · Accepted Answer

Let us say initial height is h .Velocity of ball just before hitting floor, $$u=2ghVelocity$$ of ball just after hitting floor, $$V=euLet$$ us say new height achieved is h'.So,$$V=2gh$$'⇒$$eu=2gh$$'⇒$$e2gh=2gh$$'Given : h'$$=81100h$$⇒$$e=0.9Total$$ distance travelled by $$ball=S=h+2e2h+2e4h+2e6h$$...$$=h1+e21-e2=47.632$$ mTotal time taken $$=$$ $$t=2hg+22e2hg+22e4hg+22e6hg=2hg1+e1-e=19$$ sAverage Speed , $$Vavg=$$ Total distance  Total time $$=St$$ ⇒$$Vavg=2.5$$ ms−$$1$$

A rubber ball is released from a height of 5 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 $g = 10 {ms}^{- 2}$ )

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A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to 81100 of the height through which it falls. Find the average speed of the ball. (Take g=10 ms−2 )

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A rubber ball is released from a height of 5 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 $g = 10 {ms}^{- 2}$ )