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MHT CET 2021 (21 Sep, Shift 2) · Motion Under Gravity

Motion Under Gravity question from MHT CET 2021 (21 Sep, Shift 2)

Q30MHT CET 2021 · 21 SEP · SHIFT 2Easy

Motion Under Gravity

A body at rest falls through a height 'h' with velocity 'V'. If it has to fall down further for its velocity to become three times, the distance travelled in that interval is

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Answer: (a)

  1. From rest, v2=2gh⇒h∝v2v^2 = 2gh \Rightarrow h \propto v^2.
  2. Speed 3V3V needs total fall 9h9h.
  3. Further distance =9h−h=8h= 9h - h = 8h
  4. So the answer is (a) 8 h.
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