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Physics · Motion in Plane

JEE Main 2025 — 7 April, Morning Shift — Question 59

Two projectiles are fired from ground with same initial speeds from same point at angles (45∘+α)\left(45^{\circ}+\alpha\right) and (45∘−α)\left(45^{\circ}-\alpha\right) with horizontal direction. The ratio of their times of flights is

  1. Option A:

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

  2. Option B:

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

    Correct
  3. Option C:

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

  4. Option D:

    1

Answer: B

Step-by-step solution

Time of flight for 1st 1^{\text {st }} projectile T1=24sin⁡(45+α)gT_{1}=\frac{24 \sin (45+\alpha)}{g} And

T2=24sin⁡(45−α)gT_{2}=\frac{24 \sin (45-\alpha)}{g}

So, T1T2=sin⁡(45+α)sin⁡(45−α)=cos⁡α2+sin⁡α2cos⁡α2−sin⁡α2\frac{T_{1}}{T_{2}}=\frac{\sin (45+\alpha)}{\sin (45-\alpha)}=\frac{\frac{\cos \alpha}{\sqrt{2}}+\frac{\sin \alpha}{\sqrt{2}}}{\frac{\cos \alpha}{\sqrt{2}}-\frac{\sin \alpha}{\sqrt{2}}}

⇒T1T2=1+tan⁡α1−tan⁡α\Rightarrow \frac{T_{1}}{T_{2}}=\frac{1+\tan \alpha}{1-\tan \alpha}

Answer key and solution verified before publishing.

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