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MHT CET 2025 (23 Apr, Shift 2) · Time of Flight, Maximum Height & Range

Time of Flight, Maximum Height & Range question from MHT CET 2025 (23 Apr, Shift 2)

Q6MHT CET 2025 · 23 APR · SHIFT 2Medium

Time of Flight, Maximum Height & Range

A body when projected at an angle ' θ\theta ' with the horizontal reaches a maximum height ' HH '. The time of flight of the body will be ( g=g = acceleration due to gravity)

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Answer: (c)

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