QJEEM22Ph4b33: JEE Main 2022 Physics - Class 11 JEEM22Ph4b33 Solution: Two projectiles are thrown with same initial velocity making an …

Here's how to determine the ratio of the ranges for the two projectiles:

Understanding the Concepts

• Projectile Motion: Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.


• Range: The range of a projectile is the horizontal distance it covers from its launch point to the point where it returns to the same vertical height.

Key Formula

The formula for the range (R) of a projectile is:

$R = \frac{u^2 \sin 2\theta}{g}$

where:

• R = Range


• u = Initial velocity (same for both projectiles)


• θ = Angle of projection


• g = Acceleration due to gravity (constant)

Calculations

1. Projectile 1 (45° angle):

Let the range of the projectile launched at 45° be R₁.

$R_1 = \frac{u^2 \sin (2 \times 45^{\circ})}{g} = \frac{u^2 \sin 90^{\circ}}{g} = \frac{u^2}{g}$

1. Projectile 2 (30° angle):

Let the range of the projectile launched at 30° be R₂.

$R_2 = \frac{u^2 \sin (2 \times 30^{\circ})}{g} = \frac{u^2 \sin 60^{\circ}}{g} = \frac{u^2 \sqrt{3}}{2g}$

1. Ratio of Ranges (R₁ : R₂):

Divide the range of projectile 1 by the range of projectile 2:

$\frac{R_1}{R_2} = \frac{\frac{u^2}{g}}{\frac{u^2 \sqrt{3}}{2g}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

To simplify the ratio, multiply both numerator and denominator by √3:

$\frac{R_1}{R_2} = \frac{2\sqrt{3} \times \sqrt{3}}{3 \times \sqrt{3}} = \frac{6}{3\sqrt{3}} = \frac{2}{\sqrt{3}}$

Therefore, the ratio of the ranges of the two projectiles is 2 : √3, which corresponds to Option C.