A ball is dropped from the top of a 100 m high tower on a planet.


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



Let the ball takes time t to reach the ground

Using, $S=u t+\frac{1}{2} g t^{2}$

$\Rightarrow S=0 \times t+\frac{1}{2} g t^{2}$

$\Rightarrow \quad 200=g t^{2}$

$[\because 2 S=100 \mathrm{~m}]$

$\Rightarrow t=\sqrt{\frac{200}{g}}$                  $\ldots$ (i)

In last $\frac{1}{2} s$, body travels a distance of $19 \mathrm{~m}$, so in

$\left(t-\frac{1}{2}\right)$ distance travelled $=81$

Now, $\frac{1}{2} g\left(t-\frac{1}{2}\right)^{2}=81$

$\therefore g\left(t-\frac{1}{2}\right)^{2}=81 \times 2$

$\Rightarrow\left(t-\frac{1}{2}\right)=\sqrt{\frac{81 \times 2}{g}}$

$\therefore \frac{1}{2}=\frac{1}{\sqrt{g}}(\sqrt{200}-\sqrt{81 \times 2})$

$\Rightarrow \sqrt{g}=2(10 \sqrt{2}-9 \sqrt{2})$         using (i)

$\Rightarrow \sqrt{g}=2 \sqrt{2}$

$\therefore g=8 \mathrm{~m} / \mathrm{s}^{2}$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now