A rubber ball is dropped from a height of 5 m on a planet where the acceleration due to gravity is not known, On bouncing it rises to 1 .8 m. The ball loses its velocity on bouncing by a factor of

When the ball strikes the surface of planet, the potential energy is changed into kinetic energy.
Thus
$\frac{1}{2}{\mathrm{mv}}^2=\mathrm{mgh}$
or v² =2gh=2 x g x 5=10g
After rebounce, v² = u² - 2 g h
or u² =v² + 2gh
$\therefore {\mathrm{u}}^2=0+2\times \mathrm{g}\times 1\cdot 8=3\cdot 6\mathrm{g}$
Now $\frac{{\mathrm{v}}^2}{{\mathrm{u}}^2}=\frac{10\mathrm{g}}{3\cdot 6\mathrm{g}}=\frac{100}{36} \mathrm{or} \frac{\mathrm{v}}{\mathrm{u}}=\frac{10}{6}=\frac{5}{3}$
or $\mathrm{u}=\frac{3}{5}\mathrm{v}$
Thus the ball loses it velocity by a factor of
$1-\frac{3}{5}=\frac{2}{5}$