Physics · Motion in Plane

JEE Main 2025 — 3 April, Evening Shift — Question 49

A particle is projected with velocity $u$ so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as $\frac{n u^{2}}{25 g}$ , where value of $n$ is : (Given, ' $g$ ' is the acceleration due to gravity.)

Option A:
18
Option B:
6
Option C:
12
Option D:
24
Correct

Answer: D

Step-by-step solution

$R=3 H$ $\frac{u^{2} \sin 2 \theta}{g}=\frac{3 u^{2} \sin ^{2} \theta}{2 g}$

$2 \sin \theta \cos \theta=\frac{3}{2} \sin \theta \sin \theta$

$\frac{4}{3}=\tan \theta$

$\theta=53^{\circ}$

$R=\frac{u^{2}}{g} \times 2 \times \frac{4}{5} \times \frac{3}{5}=\frac{24}{25} \frac{u^{2}}{g}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 3 April, Evening Shift
Subject: Physics
Chapter: Motion in Plane
Topic: Oblique and Horizontal Projectile Motion