The two projectiles are projected with the same initial velocities at the 15° and 30° with respect to the horizontal. The ratio of their ranges is 1:x. The value of x is

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q31 – JEE-MAIN 2026

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Home Physics Motion in a PlaneQuestion #302

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

Clue 1 of 3: Projectile Motion Range

Recall the formula for the horizontal range of a projectile.

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Step 1: Identify the formula for range✦ Active

The horizontal range of a projectile is given by the formula:

R = \frac{u^{2} \sin (2 θ)}{g}

where $u$ is the initial velocity, $θ$ is the projection angle with the horizontal, and $g$ is the acceleration due to gravity. In this problem, $u$ and $g$ are constant for both projectiles.

Step 2: Calculate ranges for given angles○ Expand

For the first projectile, the angle is $θ_{1} = 15^{\circ}$ . Its range $R_{1}$ is:

R_{1} = \frac{u^{2} \sin (2 \times 15^{\circ})}{g} = \frac{u^{2} \sin (30^{\circ})}{g}

For the second projectile, the angle is $θ_{2} = 30^{\circ}$ . Its range $R_{2}$ is:

R_{2} = \frac{u^{2} \sin (2 \times 30^{\circ})}{g} = \frac{u^{2} \sin (60^{\circ})}{g}

Step 3: Determine the ratio of ranges and find x○ Expand

The ratio of their ranges $R_{1} : R_{2}$ is:

\frac{R_{1}}{R_{2}} = \frac{\frac{u^{2} \sin (30^{\circ})}{g}}{\frac{u^{2} \sin (60^{\circ})}{g}} = \frac{\sin (30^{\circ})}{\sin (60^{\circ})}

Substitute the known values for $\sin (30^{\circ}) = \frac{1}{2}$ and $\sin (60^{\circ}) = \frac{\sqrt{3}}{2}$ :

\frac{R_{1}}{R_{2}} = \frac{1 / 2}{\sqrt{3} / 2} = \frac{1}{\sqrt{3}}

The problem states that the ratio of their ranges is $1 : x$ . Therefore, we have:

\frac{1}{x} = \frac{1}{\sqrt{3}}

Comparing both sides, we find $x = \sqrt{3}$ .

💡 Teacher's Secret Hint

Remember that $\sin (2 θ)$ is symmetric around $45^{\circ}$ , meaning $\sin (2 θ) = \sin (2 (90^{\circ} - θ))$ . However, the angles here are not complementary in that sense for the $2 θ$ term.

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Physics Question 31 – JEE-MAIN 2026

Clue 1 of 3: Projectile Motion Range

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