A ball having kinetic energy KE , is projected at an angle of 60 ° from the horizontal. What will be the kinetic energy of ball at the highest point of its flight?

Physics · Motion in Plane

JEE Main 2025 — 23 January, Evening Shift — Question 48

A ball having kinetic energy KE , is projected at an angle of $60^{\circ}$ from the horizontal. What will be the kinetic energy of ball at the highest point of its flight?

Option A:
$\frac{(\mathrm{KE})}{8}$
Option B:
$\frac{(\mathrm{KE})}{4}$
Correct
Option C:
$\frac{(\mathrm{KE})}{16}$
Option D:
$\frac{(\mathrm{KE})}{2}$

Answer: B

Step-by-step solution

Initial K.E,

K.E. $=\frac{1}{2} \mathrm{mu}^{2}$

Speed at heighest point

$\mathrm{V}=\mathrm{u} \cos 60^{\circ}=\frac{\mathrm{u}}{2}$

$\therefore \quad \mathrm{KE}_{2}=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{u}}{2}\right)^{2}$

$=\frac{1}{4} \times \frac{1}{2} \mathrm{mu}^{2}$

$=\frac{\mathrm{KE}}{4}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Evening Shift
Subject: Physics
Chapter: Motion in Plane
Topic: Oblique and Horizontal Projectile Motion