A particle is dropped under gravity from rest from a height h (g = 9.8 m/${\sec }^2$) and it travels a distance (9h/25) in the last second, the height h is

Let h distance is covered in n s.

$\Rightarrow \mathrm{h} = \frac{1}{2}{\mathrm{gn}}^2 -----\left(\mathrm{i}\right)$

Distance covered in nth second = $\frac{1}{2}\mathrm{g}(2\mathrm{n}-1)$

$\Rightarrow \frac{9\mathrm{h}}{25 } = \frac{\mathrm{g}}{2}(2\mathrm{n}-1)-----\left(\mathrm{ii}\right)$

From (i) and (ii),n = 5 Sec

so, the total time of travel is 5 sec.

Substituting in  (i), we get  h = 122.5 m