Q 11-02-116JEE MainJEE Main 2020 (8 Jan, Shift 2)Medium
A ball is dropped from the top of a $100$ m high tower on a planet. In the last $\dfrac{1}{2}$ s 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 ______.
Numerical value type. Enter your answer.
Answer: 8
Let the total fall time be $T$. In time $T - \tfrac{1}{2}$ the ball falls $100 - 19 = 81$ m:
$$\frac{1}{2}gT^2 = 100,\qquad \frac{1}{2}g\left(T - \tfrac{1}{2}\right)^2 = 81$$
Dividing and taking square roots: $\dfrac{T}{T - 1/2} = \dfrac{10}{9}$, so $T = 5$ s.
$$g = \frac{2\times100}{25} = 8\ \text{m s}^{-2}$$
Solution by Sreeraj P, M.Sc Physics