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JEE Main 2024 (27 Jan, Shift 2) · Motion Under Gravity

Motion Under Gravity question from JEE Main 2024 (27 Jan, Shift 2)

Q12JEE Main 2024 · 27 JAN · SHIFT 2Medium

Motion Under Gravity

A body falling under gravity covers two points AA and BB separated by 80 m80\text{ m} in 2 s2\text{ s}. The distance of upper point A from the starting point is _________ m\text{m} (use g=10 ms−2g = 10\text{ ms}^{-2} ).

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

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Answer: 45

  1. From A to B: 80=vA(2)+12(10)(2)2=2vA+20⇒vA=3080 = v_A(2) + \frac{1}{2}(10)(2)^2 = 2v_A + 20 \Rightarrow v_A = 30 m/s
  2. Distance of A from start: s=vA22g=90020=45s = \frac{v_A^2}{2g} = \frac{900}{20} = 45 m
  3. So the answer is 45 m.
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