Question

An object is dropped from a height $h$ from the ground. Every time it hits the ground it looses $50\%$ of its kinetic energy. The total distance covered at $t\to \infty$ is

• A. $3h$
• B. $\infty$
• C. $\frac{5}{3}h$
• D. $\frac{8}{3}h$

Answer 1

The correct option is A $3h$

By conservation of energy

$mgh=\frac{1}{2}m{v}^2$

$mg{h}_1=\frac{1}{4}m{v}^2$

$h_1=\frac{1}{2}h$

Similarly

$h_2=\frac{1}{4}h$ and so on

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

Total distance=$h+2(h_1+h_2+h_3+...)=h+2(\frac{h}{2}+\frac{h}{4}+\frac{h}{8}+...)=h+\frac{\frac{h}{2}}{1-\frac{1}{2}}=h+2h=3h$