JEE PYQ: Motion in a Plane - Question ID c9336b0be44e (JEE Main 2023)

Core Formula & Concept:

In projectile motion, the range ($R$) of a projectile is the horizontal distance it covers before returning to the same vertical level from which it was launched. The key formula for the range of a projectile projected with an initial velocity $u$ at an angle $\theta$ with the horizontal is:

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

where:

• $u$ is the initial velocity of projection,
• $\theta$ is the angle of projection with the horizontal,
• $g$ is the acceleration due to gravity.

This formula arises from combining the horizontal and vertical components of motion and using the fact that the time of flight depends on the vertical motion.

A crucial observation is that the range depends on $\sin(2\theta)$. Since $\sin(2\theta)$ is symmetric about $\theta = 45^\circ$, two angles $\theta$ and $(90^\circ - \theta)$ yield the same range. Also, the maximum range occurs at $\theta = 45^\circ$, where $\sin(90^\circ) = 1$.

--- Step-by-Step Derivation:

Given:

• Range at $\theta_1 = 15^\circ$, $R_1 = 50\text{ m}$
• We need to find range at $\theta_2 = 45^\circ$, $R_2 = ?$
• Same initial velocity $u$ is used in both cases.

Step 1: Write the range formula for both angles
For $\theta_1 = 15^\circ$: $R_1 = \frac{u^2 \sin(2 \times 15^\circ)}{g} = \frac{u^2 \sin(30^\circ)}{g}$ We know $\sin(30^\circ) = \frac{1}{2}$, so: $50 = \frac{u^2 \cdot \frac{1}{2}}{g} \implies 50 = \frac{u^2}{2g} \quad \text{(Equation 1)}$

Step 2: Write the range formula for $\theta_2 = 45^\circ$
For $\theta_2 = 45^\circ$: $R_2 = \frac{u^2 \sin(2 \times 45^\circ)}{g} = \frac{u^2 \sin(90^\circ)}{g}$ Since $\sin(90^\circ) = 1$, we have: $R_2 = \frac{u^2}{g} \quad \text{(Equation 2)}$

Step 3: Relate $R_2$ to $R_1$
From Equation 1: $50 = \frac{u^2}{2g} \implies \frac{u^2}{g} = 100$ But from Equation 2, $R_2 = \frac{u^2}{g}$, so: $R_2 = 100\text{ m}$

Conclusion: The range when the projectile is launched at $45^\circ$ is 100 m, which corresponds to option B.

--- Common Traps & Exam Tip:

Trap 1: Misremembering the range formula
Some students confuse the range formula with $R = \frac{u^2 \sin \theta}{g}$ or $R = \frac{u^2 \cos \theta}{g}$. The correct formula involves $\sin(2\theta)$, not $\sin \theta$ or $\cos \theta$. Always recall that range depends on the product of horizontal and vertical components, leading to $\sin(2\theta)$.

Trap 2: Assuming range doubles when angle doubles
Students may think that since $45^\circ$ is three times $15^\circ$, the range might scale linearly. But range depends on $\sin(2\theta)$, which is a nonlinear function. The key is to use the known range at one angle to find $u^2/g$, then apply it to the new angle.

Trap 3: Forgetting that $\sin(30^\circ) = 0.5$
A common arithmetic mistake is miscalculating $\sin(30^\circ)$. Always verify trigonometric values of standard angles.

Exam Tip:
When two ranges are given or asked at different angles with the same speed, always use the ratio: $\frac{R_1}{R_2} = \frac{\sin(2\theta_1)}{\sin(2\theta_2)}$ This avoids explicitly calculating $u$ and $g$, saving time and reducing errors.