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MHT CET 2021 (20 Sep, Shift 2) · Time of Flight, Maximum Height & Range

Time of Flight, Maximum Height & Range question from MHT CET 2021 (20 Sep, Shift 2)

Q24MHT CET 2021 · 20 SEP · SHIFT 2Easy

Time of Flight, Maximum Height & Range

A cricket player hit a ball like a projectile but the fielder caught the ball after 2 second. The maximum height reached by a ball is (g = 10 m/s2)

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Answer: (b)

  1. T=2usin⁡θgT = \frac{2u\sin\theta}{g} and H=u2sin⁡2θ2gH = \frac{u^2\sin^2\theta}{2g} give H=gT28H = \frac{gT^2}{8}.
  2. H=10×228=5 mH = \frac{10 \times 2^2}{8} = 5\text{ m}
  3. So the answer is (b) 5 m.
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