JEE PYQ: Motion in a Plane - Question ID f33c66fe429b (JEE Main 2023)

Core Formula & Concept:

In projectile motion, the range (horizontal distance traveled before landing) is a key quantity. For a projectile launched from and landing on the same horizontal plane, the range \( R \) is given by:

$R = \frac{u^2 \sin(2\theta)}{g}$ where:

• $u$ = initial speed of the projectile,
• $\theta$ = launch angle with the horizontal,
• $g$ = acceleration due to gravity.

This formula arises from combining the horizontal and vertical components of motion, ensuring the projectile returns to the same vertical level.

Step-by-Step Derivation:

Let’s compute the ranges for projectiles A and B separately.

Projectile A:

• Initial speed: $u_A = 40 \, \text{m/s}$
• Launch angle: $\theta_A = 30^\circ$

$R_A = \frac{u_A^2 \sin(2\theta_A)}{g} = \frac{40^2 \cdot \sin(60^\circ)}{10} = \frac{1600 \cdot \frac{\sqrt{3}}{2}}{10} = \frac{800\sqrt{3}}{10} = 80\sqrt{3} \, \text{m}$ Projectile B:

• Initial speed: $u_B = 60 \, \text{m/s}$
• Launch angle: $\theta_B = 60^\circ$

$R_B = \frac{u_B^2 \sin(2\theta_B)}{g} = \frac{60^2 \cdot \sin(120^\circ)}{10} = \frac{3600 \cdot \frac{\sqrt{3}}{2}}{10} = \frac{1800\sqrt{3}}{10} = 180\sqrt{3} \, \text{m}$ Ratio of Ranges: $\frac{R_A}{R_B} = \frac{80\sqrt{3}}{180\sqrt{3}} = \frac{80}{180} = \frac{4}{9}$ Thus, the ratio of their ranges is $4:9$, matching option A. Common Traps & Exam Tip:

Students often confuse the formula for range, mistakenly using $\sin\theta$ or $\cos\theta$ instead of $\sin(2\theta)$. Another frequent error is mixing up the angles or speeds when substituting values. Always double-check the trigonometric identity $\sin(2\theta) = 2\sin\theta\cos\theta$ and ensure the correct angle is used in the formula.