Q56 / 75 · JEE Main 2026 (04 April Shift 1) · Physics
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:
- A$\sqrt{2}$
- Correct answerB$\sqrt{3}$
- C$2\sqrt{3}$
- D$\dfrac{1}{\sqrt{2}}$
Solution
The horizontal range of a projectile is given by the formula $R = \dfrac{u^2 \sin(2\theta)}{g}$.
For the first projectile, the angle of projection is $\theta_1 = 15^\circ$.
$R_1 = \dfrac{u^2 \sin(2 \times 15^\circ)}{g} = \dfrac{u^2 \sin(30^\circ)}{g} = \dfrac{u^2}{2g}$
For the second projectile, the angle of projection is $\theta_2 = 30^\circ$.
$R_2 = \dfrac{u^2 \sin(2 \times 30^\circ)}{g} = \dfrac{u^2 \sin(60^\circ)}{g} = \dfrac{\sqrt{3}u^2}{2g}$
The ratio of their ranges is:
$\dfrac{R_1}{R_2} = \dfrac{\dfrac{u^2}{2g}}{\dfrac{\sqrt{3}u^2}{2g}} = \dfrac{1}{\sqrt{3}}$
Given that the ratio is $1:x$, we get $x = \sqrt{3}$.
Answer: $\sqrt{3}$
For the first projectile, the angle of projection is $\theta_1 = 15^\circ$.
$R_1 = \dfrac{u^2 \sin(2 \times 15^\circ)}{g} = \dfrac{u^2 \sin(30^\circ)}{g} = \dfrac{u^2}{2g}$
For the second projectile, the angle of projection is $\theta_2 = 30^\circ$.
$R_2 = \dfrac{u^2 \sin(2 \times 30^\circ)}{g} = \dfrac{u^2 \sin(60^\circ)}{g} = \dfrac{\sqrt{3}u^2}{2g}$
The ratio of their ranges is:
$\dfrac{R_1}{R_2} = \dfrac{\dfrac{u^2}{2g}}{\dfrac{\sqrt{3}u^2}{2g}} = \dfrac{1}{\sqrt{3}}$
Given that the ratio is $1:x$, we get $x = \sqrt{3}$.
Answer: $\sqrt{3}$
Hence, the correct answer is option: 'B'.