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Q.
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
Motion in a Straight Line
Solution:
Let $h$ distance is covered in $n\, s$.
$\Rightarrow h = \frac{1}{2} gn^2\,\,...(i)$
Distance covered in $n^{th}$ second $ = \frac{1}{2} g(2n -1)$
$ \Rightarrow \frac{9h}{25} = \frac{g}{2} (2n-1)\,\,...(ii)$
From $(i)$ and $(ii), h = 122.5 \,m$