Question

A body falls from a height h on a horizontal surface and rebounds. Then it falls again and again rebounds and so on. If the restitution coefficients is $\frac{1}{3}$, the total distance covered by the body before it comes to rest is

• A. $\frac{h}{4}$
• B. $\frac{5h}{4}$
• C. $2h$
• D. $3h$

Answer 1

Answer: (B)

$\frac{5h}{4}$

$\text{Explanation:-}$

Total Distance $=h+2h_1+2h_2+2h_3+....$

$=h+\frac{2e^2{v}^2}{2g}+\frac{2e^4{v}^2}{2g}+......$

$=\frac{v^2}{2g}+\frac{2e^2{v}^2}{2g}[1+e^2+e^4+....]$

$=\frac{v^2}{2g}+\frac{2e^2{v}^2}{2g}\left[\frac{1}{1-e^2}\right]$

$=\frac{v^2}{2g}\left[\frac{1-e^2+2e^2}{1-e^2}\right]$

$=\frac{v^2}{2g}\left[\frac{1+e^2}{1-e^2}\right]$

$=h\left[\frac{1+\frac{1}{9}}{1-\frac{1}{9}}\right]$

$=h\times \left[\frac{\frac{10}{9}}{\frac{8}{9}}\right]$

$=\frac{5h}{4}$

$\text{Hence the correct option is B}$