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MHT CET 2024 (15 May, Shift 2) · Motion Under Gravity

Motion Under Gravity question from MHT CET 2024 (15 May, Shift 2)

Q11MHT CET 2024 · 15 MAY · SHIFT 2Easy

Motion Under Gravity

A ball is released from the top of a tower of height Hm. It takes T second to reach the ground. The height of the ball from the ground after T4\frac{T}{4} second is

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Answer: (b)

  1. Free fall from rest: h=12gt2⇒h∝t2h = \frac{1}{2}gt^2 \Rightarrow h \propto t^2, with H=12gT2H = \frac{1}{2}gT^2.
  2. Fallen in T4\frac{T}{4}: H16\frac{H}{16}
  3. Height above ground =H−H16=15H16= H - \frac{H}{16} = \frac{15H}{16}
  4. So the answer is (b) 15H16\frac{15H}{16}.
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