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MHT CET 2026 (14 May, Shift 2) · Time of Flight, Maximum Height & Range

Time of Flight, Maximum Height & Range question from MHT CET 2026 (14 May, Shift 2)

Q1MHT CET 2026 · 14 MAY · SHIFT 2Easy

Time of Flight, Maximum Height & Range

If the initial velocity of projection be doubled, keeping the angle of projection same, the maximum height reached by it will

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Answer: (c)

  1. H=u2sin⁡2θ2g⇒H∝u2H = \frac{u^2\sin^2\theta}{2g} \Rightarrow H \propto u^2 for the same angle.
  2. Doubling uu gives H′=22H=4HH' = 2^2 H = 4H
  3. So the answer is (c) become four times.
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