A particle is dropped under gravity from rest from a height h(g = 9.8 m/sec²) and it travels a distance 9h/25 in the last second, the height h is:

Let h distance is covered in n sec
$\Rightarrow \mathrm{h}=\frac{1}{2}{\mathrm{gn}}^2 .....\left(\mathrm{i}\right)$
Distance covered in ${\mathrm{n}}^{\text{th }}\sec =\frac{1}{2}\mathrm{g}(2\mathrm{n}-1)$
$\Rightarrow \frac{9\mathrm{h}}{25}=\frac{\mathrm{g}}{2}(2\mathrm{n}-1) .....\left(\mathrm{ii}\right)$
From (i) and (ii), h=122.5 m