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MHT CET 2026 (17 Apr, Shift 1) · Comparing Two Projectiles

Comparing Two Projectiles question from MHT CET 2026 (17 Apr, Shift 1)

Q6MHT CET 2026 · 17 APR · SHIFT 1Medium

Comparing Two Projectiles

Two balls A and B are projected at an angle of 45∘45^\circ and 60∘60^\circ respectively, so that the maximum heights reached are same for both. The ratio of initial velocity of projection of ball A to that for ball B is

(sin⁡30∘=cos⁡60∘=12, sin⁡45∘=cos⁡45∘=12, sin⁡60∘=cos⁡30∘=32)\left(\sin 30^\circ = \cos 60^\circ = \frac{1}{2},\ \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}},\ \sin 60^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}\right)

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Answer: (d)

  1. Equal H=u2sin⁡2θ2gH = \frac{u^2\sin^2\theta}{2g} means uAsin⁡45∘=uBsin⁡60∘u_A\sin 45^\circ = u_B\sin 60^\circ.
  2. uAuB=sin⁡60∘sin⁡45∘=3/21/2=32\frac{u_A}{u_B} = \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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