A particle is dropped under gravity from rest from a height h (g = 9.8 m/s² ) and it travels a distance 9h/25 in the last second, the height h is

Let h distance is covered in t second

$\Rightarrow h=\frac{1}{2}g{t}^2$

Distance covered in t^th second $=\frac{1}{2}g(2t-1)$

$$\begin{gathered} \Rightarrow \frac{9h}{25}=\frac{g}{2}(2t-1) \\ \frac{9}{25}\frac{1}{2}g{t}^2=\frac{g}{2}(2t-1) \\ \frac{(2t-1)}{t^2}=\frac{9}{25} \\ t=5 sec \end{gathered}$$

then,  h=122.5 m