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Physics · Motion in Plane

JEE Main 2025 — 8 April, Evening Shift — Question 49

Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball. T1T_{1} and T2T_{2} are the total flying times of first and second ball, respectively, then the ratio of T1T_{1} and T2T_{2} is

  1. Option A:

    22:12 \sqrt{2}: 1

    Correct
  2. Option B:

    2:1\sqrt{2}: 1

  3. Option C:

    4:14: 1

  4. Option D:

    2:12: 1

Answer: A

Step-by-step solution

H1=u2sin⁡2θ12gH_{1}=\frac{u^{2} \sin ^{2} \theta_{1}}{2 g} H2=u2sin⁡2θ22gH_{2}=\frac{u^{2} \sin ^{2} \theta_{2}}{2 g}

T1=2usin⁡θ1gT_{1}=\frac{2 u \sin \theta_{1}}{g} T2=2usin⁡θ2gT_{2}=\frac{2 u \sin \theta_{2}}{g}

H1H2=sin⁡2θ1sin⁡2θ2\frac{H_{1}}{H_{2}}=\frac{\sin ^{2} \theta_{1}}{\sin ^{2} \theta_{2}}

T1T2=sin⁡θ1sin⁡θ2\frac{T_{1}}{T_{2}}=\frac{\sin \theta_{1}}{\sin \theta_{2}}

8=(T1T2)28=\left(\frac{T_{1}}{T_{2}}\right)^{2}

⇒T1T2=22\Rightarrow \frac{T_{1}}{T_{2}}=2 \sqrt{2}

T1:T2=22:1T_{1}: T_{2}=2 \sqrt{2}: 1

Answer key and solution verified before publishing.

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