Question

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

• A. 100 m
• B. 122.5 m
• C. 145 m
• D. 167.5 m

Answer 1

The correct option is B 122.5 m

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