Physics · Motion in Plane

JEE Main 2024 — 1 February, Shift 1 — Question 44

A particle moving in a circle of radius R with uniform speed takes time T to complete one revolution. If this particle is projected with the same speed at an angle $\theta$ to the horizontal, the maximum height attained by it is equal to $4 R$ . The angle of projection $\theta$ is then given by :

Option A:
${{\sin }^{-1}}{{\left[ \frac{2g{{T}^{2}}}{{{\pi }^{2}}R} \right]}^{\frac{1}{2}}}$
Correct
Option B:
${{\sin }^{-1}}{{\left[ \frac{{{\pi }^{2}}R}{2g{{T}^{2}}} \right]}^{\frac{1}{2}}}$
Option C:
$\cos ^{-1}\left[\frac{2 g T^{2}}{\pi^{2} R}\right]^{\frac{1}{2}}$
Option D:
$\cos ^{-1}\left[\frac{\pi \mathrm{R}}{2 \mathrm{gT}^{2}}\right]^{\frac{1}{2}}$

Answer: A

Step-by-step solution

$\frac{2\pi \text{R}}{\text{T}}=\text{V}$

             $\text{ }\!\!~\!\!\text{ Maximum }\!\!~\!\!\text{ height }\!\!~\!\!\text{ H}=\frac{{{\text{v}}^{2}}\text{si}{{\text{n}}^{2}}\theta }{2\text{ }\!\!~\!\!\text{ g}}$

$4R=\frac{4{{\pi }^{2}}{{R}^{2}}}{{{T}^{2}}2g}\text{si}{{\text{n}}^{2}}\theta $

$\begin{array}{*{35}{r}}{} & \text{sin}\theta =\sqrt{\frac{2\text{g}{{\text{T}}^{2}}}{{{\pi }^{2}}\text{R}}} \\{} & \theta =\text{si}{{\text{n}}^{-1}}{{\left( \frac{2\text{g}{{\text{T}}^{2}}}{{{\pi }^{2}}\text{R}} \right)}^{\frac{1}{2}}} \\\end{array}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 1
Subject: Physics
Chapter: Motion in Plane
Topic: Oblique and Horizontal Projectile Motion