Q 11-03-048JEE MainJEE Main 2026 (4 Apr, Shift 1)Easy
The two projectiles are projected with the same initial velocities at the $15^\circ$ and $30^\circ$ with respect to the horizontal. The ratio of their ranges is $1 : x$. The value of $x$ is
Answer: (B) $\sqrt{3}$
Concept. For the same speed, $R=\dfrac{u^2\sin2\theta}{g}\propto\sin2\theta$.
Step 1. Ratio.
$$\frac{R_{15}}{R_{30}}=\frac{\sin30^\circ}{\sin60^\circ}=\frac{1/2}{\sqrt3/2}=\frac1{\sqrt3}$$
Answer. $1:\sqrt3$, so $x=\sqrt3$ (Option B).
Solution by Sreeraj P, M.Sc Physics
