Physics · Motion in Plane

JEE Main 2025 — 24 January, Morning Shift — Question 54

An object of mass ' $m$ ' is projected from origin in a vertical xy plane at an angle $45^{\circ}$ with the x -axis with an initial velocity $\mathrm{v}_{0}$ . The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [ g is acceleration due to gravity]

Option A:
$\frac{\mathrm{mv}_{0}^{3}}{2 \sqrt{2} g}$ along negative $z$ -axis
Option B:
$\frac{m v_{0}^{3}}{2 \sqrt{2} g}$ along positive $z$ -axis
Option C:
$\frac{\mathrm{mv}_{0}^{3}}{4 \sqrt{2} g}$ along positive z -axis
Option D:
$\frac{m v_{0}^{3}}{4 \sqrt{2} g}$ along negative $z$ -axis
Correct

Answer: D

Step-by-step solution

$H=\frac{\left(\frac{\mathrm{v}_{0}}{\sqrt{2}}\right)^{2}}{2 \mathrm{~g}}=\frac{\mathrm{v}_{0}^{2}}{4 \mathrm{~g}}$

$\mathrm{L}=\mathrm{mvh}$

$\mathrm{L}=\mathrm{m} \frac{\mathrm{v}_{0}}{\sqrt{2}} \frac{\mathrm{v}_{0}^{2}}{4 \mathrm{~g}}$

Answer key and solution verified before publishing.

Practise Motion in Plane

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 24 January, Morning Shift
Subject: Physics
Chapter: Motion in Plane
Topic: Oblique and Horizontal Projectile Motion