A stone falls freely under gravity. It covers distances ${\mathrm{h}}_1,{\mathrm{h}}_2\mathrm{and}{\mathrm{h}}_3$in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between ${\mathrm{h}}_1,{\mathrm{h}}_2\mathrm{and}{\mathrm{h}}_3$ is

Distance covered by the stone in first 5 seconds

(i.e. t = 5s) is$h_1=\frac{1}{2}g{\left(5\right)}^2=\frac{25}{2}g---\left(i\right)$

Distance travelled by the stone in next 5 seconds (i.e. t = 10 s) is

$h_1+h_2=\frac{1}{2}g{\left(10\right)}^2=\frac{100}{2}g---\left(ii\right)$

Distance travelled by the stone in next 5 seconds (i.e. t =15 s) is

$h_1+h_2+h_3=\frac{1}{2}g{\left(15\right)}^2=\frac{225}{2}g---\left(iii\right)$

Subtract (i) from (ii), we get

$$\begin{gathered} \left(h_1+h_2\right)-h_1=\frac{100}{2}g-\frac{25}{2}g=\frac{75}{2}g \\ {h}_2=\frac{75}{2}g=3h_1---\left(iv\right) \end{gathered}$$

Subtract (ii) from (iii), we get

$$\begin{gathered} \left(h_1+h_2+h_3\right)-\left(h_2+h_1\right)=\frac{225}{2}g-\frac{100}{2}g \\ {h}_3=\frac{125}{2}g=5h_1---\left(v\right) \end{gathered}$$

From (i), (iv) and (v), we get

$h_1=\frac{h_2}{3}=\frac{h_3}{5}$