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/s2) near the surface on that planet is …..
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Ans: a = 8 m/s2
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/s2