A steel ball of mass $$5\,g$$  is thrown downward with velocity $$10\,m/s$$  from height $$19.5\,m.$$  It penetrates sand by $$50\,cm.$$  The change in mechanical energy will be
$$\left( {g = 10\,m/{s^2}} \right)$$

Solution :
$${v^2} = {u^2} + 2gh = {\left( {10} \right)^2} + 2 \times 10 \times 19.5 = 490$$
$$K.E.$$  at the ground $$ = \frac{1}{2}m{v^2} = \frac{1}{2} \times \frac{5}{{1000}} \times 490 = \frac{{49}}{{40}}\,J$$
$$P.E. = mgh = \frac{5}{{1000}} \times 10 \times \left( {\frac{{ - 50}}{{100}}} \right) = - \frac{1}{{40}}\,J$$
∴ Change in energy $$ = \frac{{49}}{{40}} - \left( { - \frac{1}{{40}}} \right)$$
$$ = \frac{{50}}{{40}} = 1.25\,J$$