A rubber ball is released from a height of 5, | vedclass

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(D) The initial height is $h = 5\, m$. The ball rises to a fraction $f = \frac{81}{100}$ of its previous height after each bounce. Thus,$e^2 = \frac{8

Vedclass · Accepted Answer

$2.50$

Vedclass · Answer

(D) The initial height is $h = 5\, m$. The ball rises to a fraction $f = \frac{81}{100}$ of its previous height after each bounce. Thus,$e^2 = \frac{81}{100}$,which means the coefficient of restitution $e = 0.9$.Total distance $S = h + 2(fh) + 2(f^2h) + \dots = h + 2h(f + f^2 + \dots) = h + 2h \left( \frac{f}{1-f} ight) = h \left( 1 + \frac{2f}{1-f} ight) = h \left( \frac{1+f}{1-f} ight)$.Substituting $h = 5$ and $f = 0.81$: $S = 5 \left( \frac{1.81}{0.19} ight) = 5 	imes \frac{181}{19} \approx 47.63\, m$.Total time $t = \sqrt{\frac{2h}{g}} + 2e\sqrt{\frac{2h}{g}} + 2e^2\sqrt{\frac{2h}{g}} + \dots = \sqrt{\frac{2h}{g}} \left( 1 + 2e + 2e^2 + \dots ight) = \sqrt{\frac{2h}{g}} \left( 1 + \frac{2e}{1-e} ight) = \sqrt{\frac{2h}{g}} \left( \frac{1+e}{1-e} ight)$.Substituting $h = 5, g = 10, e = 0.9$: $t = \sqrt{\frac{2 	imes 5}{10}} \left( \frac{1+0.9}{1-0.9} ight) = 1 	imes \frac{1.9}{0.1} = 19\, s$.Average speed $v_{av} = \frac{S}{t} = \frac{5 	imes 1.81 / 0.19}{19} = \frac{5 	imes 181 / 19}{19} = \frac{905}{361} \approx 2.5\, m/s$.

$A$ 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 $g = 10\, m/s^2$)

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