Q: A ball is dropped from the top of a 100 m high tower on a planet . In the last 1/2 sec before hitting the ground , it covers a distance of 19 m . Acceleration due to gravity (in m/s^{2}) near the surface on that planet is …..

**Click to See Solution : **

Ans: a = 8 m/s

^{2}Sol: Let the time taken by ball to travel 81 m is t sec

So time taken to travel 100 m is (t + 1/2) sec

$\displaystyle s = ut + \frac{1}{2} a t^2 $

$\displaystyle 81 = \frac{1}{2} a t^2 $

$\displaystyle t = 9 \sqrt{\frac{2}{a}} $ …(i)

$\displaystyle 100 = \frac{1}{2} a (t + \frac{1}{2})^2 $

$\displaystyle (t + \frac{1}{2}) = 10 \sqrt{\frac{2}{a}} $ …(ii)

$\displaystyle (9 \sqrt{\frac{2}{a}} + \frac{1}{2}) = 10 \sqrt{\frac{2}{a}} $

$\displaystyle \frac{1}{2} = \sqrt{\frac{2}{a}}$

a = 8 m/s^{2}