A stone falls freely under gravity. It covers distances $${h_1},{h_2}$$  and $${h_3}$$ in the first $$5s,$$ the next $$5s$$ and the next $$5s$$ respectively. The relation between $${h_1},{h_2}$$  and $${h_3}$$ is

Solution :
For free fall from a height, $$u = 0$$
$$\therefore $$ Distance covered by stone in first $$5 s,$$
$${h_1} = 0 + \frac{1}{2}g{\left( 5 \right)^2} = \frac{{25}}{2}g\,......\left( {\text{i}} \right)$$
$$\therefore $$ Distance covered in first $$10 s,$$
$${s_2} = 0 + \frac{1}{2}g{\left( {10} \right)^2} = \frac{{100}}{2}g$$
$$\therefore $$ Distance covered in second $$5 s$$
$${h_2} = {s_2} - {h_1} = \frac{{100}}{2}g - \frac{{25}}{2}g = \frac{{72}}{2}g\,......\left( {{\text{ii}}} \right)$$
$$\therefore $$ Distance covered in first $$15 s,$$
$${s_3} = 0 + \frac{1}{2}g{\left( {15} \right)^2} = \frac{{225}}{2}g$$
$$\therefore $$ Distance coveted in last $$5 s,$$
$${h_3} = {s_3} - {s_2} = \frac{{225}}{2}g - \frac{{100}}{2}g = \frac{{125}}{2}g\,......\left( {{\text{iii}}} \right)$$
From Eqs. (i), (ii) and (iii), we get
$$\eqalign{ & {h_1}:{h_2}:{h_3} = \frac{{25}}{2}g:\frac{{75}}{2}g:\frac{{125}}{2}g = 1:3:5 \cr & \Rightarrow {h_1} = \frac{{{h_2}}}{3} = \frac{{{h_3}}}{5} \cr} $$