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JEE Main 2022 (29 Jun, Shift 1) · Motion Under Gravity

Motion Under Gravity question from JEE Main 2022 (29 Jun, Shift 1)

Q27JEE Main 2022 · 29 JUN · SHIFT 1Medium

Motion Under Gravity

Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms−1{}^{-1}. [use g = 10 ms−2{}^{-2}]:

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Answer: (d)

  1. Meeting point is 180−100=80 m180 - 100 = 80\text{ m} below the top.
  2. Ball A: 80=12(10)t2⇒t=4 s80 = \frac{1}{2}(10)t^2 \Rightarrow t = 4\text{ s}
  3. Ball B moves for 4−2=2 s4 - 2 = 2\text{ s}: 80=2u+12(10)(2)2=2u+2080 = 2u + \frac{1}{2}(10)(2)^2 = 2u + 20
  4. u=30 m/su = 30\text{ m/s}, so the answer is (d) 30.
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