Question

A particle is dropped under gravity from rest from a height $ h(g=9.8\,m/{{s}^{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

Answer: (B)

Given, particle is dropped under gravity from rest of height $h$ and $g=9.8\, m / s ^{2}$ and travels a distance $\frac{9 h}{25}$ in the last second.
$\therefore h_{n t h} =u+\frac{1}{2} g(2 n-1) $
$=\frac{1}{2} g(2 n-1) \quad(\because u=0) $
$\Rightarrow \frac{9 h}{25} =\frac{1}{2} g(2 n-1)$
But $h=\frac{1}{2} g n^{2}$
$\Rightarrow \frac{9}{25} \times \frac{1}{2} g n^{2}=\frac{1}{2} g(2 n-1)$
$\Rightarrow 9 n^{2}-50 n+25=0$
$\Rightarrow n=5, \frac{5}{9}$
But $n=\frac{5}{9}$ is not possible.
$\therefore n=5\, s$
Then, $h=\frac{1}{2} \times 9.8 \times 5 \times 5$
$\Rightarrow h=122.5 \,m / s$