Question

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