JEE PYQ: Motion In Two Dimensions Question 14
Question 14 - 2020 (08 Jan Shift 2)
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 ms$^{-2}$) near the surface on that planet is ________.
Show Answer
Answer: 8
Solution
From $200 = gt^2$ and distance in last 0.5 s being 19 m: $\sqrt{g} = 2\sqrt{2}$, so $g = 8$ m/s$^2$.
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