Question

A ball is dropped from the top of a $100\text{m}$ high tower on a planet. In the last $\frac{1}{2}\text{s}$ before hitting the ground, it covers a distance of $19\text{m}$. Acceleration due to gravity (in ${\text{ms}}^{-2}$) near the surface on that planet is

Answer 1

Let the ball takes time to reach the ground
Using,
$S=ut+\frac{1}{2}g{t}^2$
$\Rightarrow S=0\times t+\frac{1}{2}g{t}^2$
$\Rightarrow 200=g{t}^2[\because s=100\text{m}]$
$\Rightarrow t=\sqrt{\frac{200}{g}}\dots \left(i\right)$
In the last $\frac{1}{2}\text{s}$, body travels a distance of $19\text{m}$, so in $(t-\frac{1}{2})$ distance trvalled $=81$
Now, $\frac{1}{2}g{(t-\frac{1}{2})}^2=81$
$\therefore g{(t-\frac{1}{2})}^2=81\times 2$
$\Rightarrow (t-\frac{1}{2})=\sqrt{\frac{81\times 2}{g}}$

using $\left(i\right)$
$\therefore \frac{1}{2}=\frac{1}{\sqrt{g}}(\sqrt{200}-\sqrt{81\times 2})$
$\Rightarrow \sqrt{g}=2(10\sqrt{2}-9\sqrt{2})$
$\Rightarrow \sqrt{g}=2\sqrt{2}$
$\therefore g=8{\text{ms}}^{-2}$