Explanation

(b) Potential energy=Kinetic energy
Ie, $mgh = \frac{1}{2}mv^{2}$
Or $\ \ \ \ \ \ \ \ \ \ \ \ v = \sqrt{2gh}$
If $h_{1}$ and $h_{2}$ are initial and final heights, then
$v_{1} = \sqrt{2gh_{1}},\ \ \ v_{2} = \sqrt{2gh_{2}}$
Loss in velocity
$\mathrm{\Delta}v = v_{1} - v_{2} = \sqrt{2gh_{1}} - \sqrt{2gh_{2}}$
$\therefore$Fractional loss in velocity =$\frac{\mathrm{\Delta}v}{v_{1}}$
$= \frac{\sqrt{2gh_{1}}\ - \sqrt{2gh_{2}}}{\sqrt{2gh_{1}}}$
$\frac{\mathrm{\Delta}v}{v_{1}} = 1 - \sqrt{\frac{h_{2}}{h_{1}}}$
$= 1 - \sqrt{\frac{1.8}{5}}$
$= 1 - \sqrt{0.36} = 1 - 0.6 = 0.4 = \frac{2}{5}$