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MHT CET 2026 (16 Apr, Shift 2) · Comparing Two Projectiles

Comparing Two Projectiles question from MHT CET 2026 (16 Apr, Shift 2)

Q1MHT CET 2026 · 16 APR · SHIFT 2Easy

Comparing Two Projectiles

Two objects are projected with same velocity 'u' at different angles α\alpha and β\beta with the horizontal. If α+β=90∘\alpha + \beta = 90^\circ, the ratio of horizontal range of the first object to the second object will be [sin⁡(π−θ)=sin⁡θ][\sin(\pi - \theta) = \sin\theta]

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Answer: (c)

  1. R=u2sin⁡2θgR = \frac{u^2\sin 2\theta}{g}; with β=90∘−α\beta = 90^\circ - \alpha, sin⁡2β=sin⁡(180∘−2α)=sin⁡2α\sin 2\beta = \sin(180^\circ - 2\alpha) = \sin 2\alpha.
  2. So R1=R2R_1 = R_2.
  3. So the answer is (c) 1:11 : 1.
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