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MHT CET 2026 (13 May, Shift 2) · Comparing Two Projectiles

Comparing Two Projectiles question from MHT CET 2026 (13 May, Shift 2)

Q4MHT CET 2026 · 13 MAY · SHIFT 2Medium

Comparing Two Projectiles

If two spheres are projected at angles 45∘45^\circ and 60∘60^\circ with horizontal and the maximum heights reached are same. The ratio of their initial velocities is (sin⁡45∘=12,sin⁡60∘=32\sin 45^\circ = \frac{1}{\sqrt{2}}, \sin 60^\circ = \frac{\sqrt{3}}{2})

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Answer: (d)

  1. Equal H=u2sin⁡2θ2gH = \frac{u^2\sin^2\theta}{2g} means u1sin⁡45∘=u2sin⁡60∘u_1\sin 45^\circ = u_2\sin 60^\circ.
  2. u1u2=sin⁡60∘sin⁡45∘=3/21/2=32\frac{u_1}{u_2} = \frac{\sin 60^\circ}{\sin 45^\circ} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}}
  3. So the answer is (d) 3:2\sqrt{3} : \sqrt{2}.
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