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JEE Main 2020 (8 Jan, Shift 2) · Motion Under Gravity

Motion Under Gravity question from JEE Main 2020 (8 Jan, Shift 2)

Q41JEE Main 2020 · 8 JAN · SHIFT 2Medium

Motion Under Gravity

A ball is dropped from the top of a 100 m high tower on a planet. In the last 12 s\frac{1}{2}\text{ 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 _____.

Numerical answer type. Work it out, then open the answer.

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Answer: 8

  1. Total time TT: 100=12gT2100 = \frac{1}{2}gT^2; before the last 0.5 s it fell 100−19=81100 - 19 = 81 m: 81=12g(T−0.5)281 = \frac{1}{2}g(T - 0.5)^2
  2. Divide: TT−0.5=10081=109⇒T=5\frac{T}{T - 0.5} = \sqrt{\frac{100}{81}} = \frac{10}{9} \Rightarrow T = 5 s
  3. g=2×10052=8g = \frac{2 \times 100}{5^2} = 8 m/s2^2
  4. So the answer is 8 m/s2^2.
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