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Physics · Motion in Plane

JEE Main 2025 — 29 January, Morning Shift — Question 24

Two projectiles are fired with same initial speed from same point on ground at angles of

(45∘−α)\left(45^{\circ}-\alpha\right) and (45∘+α)\left(45^{\circ}+\alpha\right),

respectively, with the horizontal direction. The ratio of their maximum heights attained is :

  1. Option A:

    1−tan⁡α1+tan⁡α\frac{1-\tan \alpha}{1+\tan \alpha}

  2. Option B:

    1+sin⁡α1−sin⁡α\frac{1+\sin \alpha}{1-\sin \alpha}

  3. Option C:

    1−sin⁡2α1+sin⁡2α\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}

    Correct
  4. Option D:

    1+sin⁡2α1−sin⁡2α\frac{1+\sin 2 \alpha}{1-\sin 2 \alpha}

Answer: C

Step-by-step solution

HMax=(usin⁡θ)22 g\mathrm{H}_{\mathrm{Max}}=\frac{(\mathrm{u} \sin \theta)^{2}}{2 \mathrm{~g}} (Hmax⁡)1(Hmax⁡)2=u2sin⁡2(45−α)u2sin⁡2(45+α)\frac{\left(\mathrm{H}_{\max }\right)_{1}}{\left(\mathrm{H}_{\max }\right)_{2}}=\frac{\mathrm{u}^{2} \sin ^{2}(45-\alpha)}{\mathrm{u}^{2} \sin ^{2}(45+\alpha)}

=(12cos⁡α−12sin⁡α)2(12cos⁡α+12sin⁡α)2=\frac{\left(\frac{1}{\sqrt{2}} \cos \alpha-\frac{1}{\sqrt{2}} \sin \alpha\right)^{2}}{\left(\frac{1}{\sqrt{2}} \cos \alpha+\frac{1}{\sqrt{2}} \sin \alpha\right)^{2}}

=1−sin⁡2α1+sin⁡2α=\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
29 January, Morning Shift
Subject
Physics
Chapter
Motion in Plane
Topic
Oblique and Horizontal Projectile Motion

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