Question

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

• A. $2.5m{s}^{-1}$
• B. $3.5m{s}^{-1}$
• C. $3.0m{s}^{-1}$
• D. $2.0m{s}^{-1}$

Answer 1

The correct option is A

$2.5m{s}^{-1}$

Step 1. Given data and diagram for the question:

Height from which the ball is dropped initially, $\left(h\right)=5m$

The ratio of the rise of the ball to the previous height $\left(\frac{h\text{'}}{h}\right)=\left(\frac{81}{100}\right)$

Acceleration due to gravity, $g=10m{s}^{-2}$

Let $d$ be the total horizontal distance covered by the ball and

Let $T$ be the total time taken by the ball to cover the horizontal distance $d$

When the ball bounces, it continuously loses energy which is ascertained by the decrease in the height to which the ball rises in the next bounce.

The loss of height is given by:

$h\text{'}=e^{2}h$

Where $e$is a constant.

Step 2. Find the horizontal distance traveled by the ball:

The total distance traveled by the ball,

$d=h+2e^{2}h+2e^{4}h+2e^{6}h+2e^{8}h+\dots$

$=1+2e^{2}h\left(1+e^2+e^4+e^6+\dots \right)$

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

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

Step 3. Find the Total time taken by the ball:

The total time taken by the ball,

$T=h+2et+2e^{2}t+2e^{3}t+2e^{4}t+\dots$

$=t+2et\left(1+e+e^2+e^3+e^4+\dots \right)$

$=1+2et\left(\frac{1}{1-e}\right)$

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

For $1sec$ we have,

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

Step 3. Find the value of $\mathit{e}$ for the case:

From the figure, $e^2=\frac{h\text{'}}{h}$

$=\frac{81}{100}$

$\Rightarrow$ $e=0.9$

Step 4. Find ${\mathit{V}}_{\mathbf{a}\mathbf{v}\mathbf{g}}$

The average speed of the ball is given by

$V_{avg}=\frac{d}{T}$

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

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

$=5\frac{\left(1+0.9^2\right)}{{\left(1+0.9\right)}^2}$

$\Rightarrow$$V_{avg}=2.5m{s}^{-1}$

Hence, option (A) is correct