Question

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

• A. $2.50{\text{m s}}^{-1}$
• B. $3.50{\text{m s}}^{-1}$
• C. $3.0{\text{m s}}^{-1}$
• D. $2.0{\text{m s}}^{-1}$

Answer 1

The correct option is A $2.50{\text{m s}}^{-1}$

Let the situation of the ball be as shown in the below figure.
Total distance: $d=h+2e^{2}h+2e^{4}h+2e^{6}h+2e^{8}h+...$

$d=h+2e^{2}h(1+e^2+e^4+e^6+...)$

$d=h+2e^{2}h\left(\frac{1}{1-e^2}\right)$

$d=\frac{(1-e^2)h+2e^{2}h}{1-e^2}=\frac{h(1+e^2)}{1-e^2}$

Total time: $t=T+2eT+2e^{2}T+2e^{3}T+...$ Where $T=\sqrt{\frac{2h}{g}}=1$

$t=T+2eT(1+e+e^2+e^3+...)$

$t=T+2eT\left(\frac{1}{1-e}\right)$

$t=\frac{T(1+e)}{1-e}$

Now, average speed of the ball
$V_{avg}=\frac{d}{t}=\frac{h\frac{(1+e^2)}{(1-e^2)}}{T\left(\frac{1+e}{1-e}\right)}$

$V_{avg}=\frac{5}{1}\left(\frac{1+e^2}{(1+e)(1-e)}\frac{(1-e)}{(1+e)}\right)$

$V_{avg}=\frac{5(1+e^2)}{(1+e{)}^2}$

$\because h^{\prime }=e^{2}h$
From the question: $\frac{81}{100}=e^2$
$e=\frac{9}{10}=0.9$

$V_{avg}=\frac{5(1+\frac{81}{100})}{(1+0.9{)}^2}$

$V_{avg}=2.50\text{m/s}$