Q71 / 75 · JEE Main 2020 (08 Jan Shift 2) · Physics

A ball is dropped from the top of a $100 m$ high tower on a planet. In the last $\frac{1}{2} s$ 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______

Correct answer: 8

Solution

$\frac{g {[t - \frac{1}{2}]}^{2}}{2} \times \frac{1}{2} = 81$ ... (1)
$\frac{1}{2} g t^{2} = 100$ …. (2)
devide equation $(1) & (2)$
$\frac{t}{t - \frac{1}{2}} = \frac{10}{9}$

$t = 5 s$

put t in equation $(2)$

$g = 8 \frac{m}{s^{2}}$

Hence, the correct answer is 8.

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Solution

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A ball is dropped from the top of a $100 m$ high tower on a planet. In the last $\frac{1}{2} s$ 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______