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MHT CET 2026 (14 May, Shift 1) · Time of Flight, Maximum Height & Range

Time of Flight, Maximum Height & Range question from MHT CET 2026 (14 May, Shift 1)

Q4MHT CET 2026 · 14 MAY · SHIFT 1Medium

Time of Flight, Maximum Height & Range

A stone is thrown at an angle θ\theta with the horizontal reaches a maximum height HH, then the time of flight of stone will be (gg = acceleration due to gravity)

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Answer: (c)

  1. H=u2sin⁡2θ2g⇒usin⁡θ=2gHH = \frac{u^2\sin^2\theta}{2g} \Rightarrow u\sin\theta = \sqrt{2gH}
  2. T=2usin⁡θg=22gHg=22HgT = \frac{2u\sin\theta}{g} = \frac{2\sqrt{2gH}}{g} = 2\sqrt{\frac{2H}{g}}
  3. So the answer is (c) 2(2Hg)1/22\left(\frac{2H}{g}\right)^{1/2}.
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