JEE Main 2026 — Apr 4, Shift 1 — Physics solution

JEE Main2026Apr 4, Shift 1Physics

The two projectiles are projected with the same initial velocities at the $1 5^{\circ}$ and $3 0^{\circ}$ with respect to the horizontal. The ratio of their ranges is $1 : x$ . The value of $x$ is

A
$2$
B
$3$
C
$23$
D
$\frac{1}{2}$

Solution

Range of a projectile: $R = \frac{u ^{2} sin 2 θ}{g}$ . Same $u$ , so ratio depends only on $sin 2 θ$ .

$\frac{R _{1}}{R _{2}} = \frac{sin ( 2 \times 1 5 ^{\circ} )}{sin ( 2 \times 3 0 ^{\circ} )} = \frac{sin 3 0 ^{\circ}}{sin 6 0 ^{\circ}} = \frac{1/2}{3 /2} = \frac{1}{3} .$

Hence $x = 3 .$

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →