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MHT CET 2026 (13 May, Shift 1) · Velocity, Momentum & Energy in Flight

Velocity, Momentum & Energy in Flight question from MHT CET 2026 (13 May, Shift 1)

Q4MHT CET 2026 · 13 MAY · SHIFT 1Medium

Velocity, Momentum & Energy in Flight

A particle is projected from ground with speed u and at an angle θ\theta with horizontal. If at maximum height from ground, the speed of a particle is half times its initial velocity of projection then its maximum height attained is (cos⁡60∘=0.5, sin⁡60∘=3/2)(\cos 60^\circ = 0.5,\ \sin 60^\circ = \sqrt{3}/2) (g = acceleration due to gravity)

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Answer: (b)

  1. Speed at the top =ucos⁡θ=u2⇒θ=60∘= u\cos\theta = \frac{u}{2} \Rightarrow \theta = 60^\circ
  2. H=u2sin⁡2θ2g=u2(3/4)2gH = \frac{u^2\sin^2\theta}{2g} = \frac{u^2 (3/4)}{2g}
  3. H=3u28gH = \frac{3u^2}{8g}
  4. So the answer is (b).
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