JEE PYQ: Motion in a Plane - Question ID 863163d4eae7 (JEE Main 2023)

In projectile motion, two key quantities are the range and the maximum height of the projectile. When two projectiles are launched with the same initial speed but at different angles, their ranges can be equal if the angles are complementary (i.e., their sum is $90^{\circ}$). This is a fundamental property derived from the range formula.

The range $R$ of a projectile launched with speed $u$ at an angle $\theta$ with the horizontal is given by: $R = \frac{u^2 \sin(2\theta)}{g}$ where $g$ is the acceleration due to gravity.

The maximum height $H$ attained by the projectile is: $H = \frac{u^2 \sin^2(\theta)}{2g}$

If two projectiles have the same range, then: $\sin(2\theta_1) = \sin(2\theta_2)$ This equation holds when $\theta_1 + \theta_2 = 90^{\circ}$ (or when $\theta_1 = \theta_2$, which is trivial and not the case here).

Step 1: Identify the given data
- Initial speed of both projectiles: $u = 40 \mathrm{~ms^{-1}}$ - One angle of projection: $\theta_1 = 60^{\circ}$ - Acceleration due to gravity: $g = 10 \mathrm{~ms^{-2}}$ - Both projectiles have the same range.

Step 2: Determine the second angle of projection
Since the ranges are equal, the angles must satisfy: $\sin(2\theta_1) = \sin(2\theta_2)$ Given $\theta_1 = 60^{\circ}$, we have: $\sin(120^{\circ}) = \sin(2\theta_2)$ We know that $\sin(120^{\circ}) = \sin(60^{\circ}) = \frac{\sqrt{3}}{2}$. The general solution for $\sin A = \sin B$ is: $A = B + 2\pi n \quad \text{or} \quad A = \pi - B + 2\pi n, \quad n \in \mathbb{Z}$ For angles in degrees, this becomes: $2\theta_2 = 120^{\circ} \quad \text{or} \quad 2\theta_2 = 180^{\circ} - 120^{\circ} = 60^{\circ}$ Thus: $\theta_2 = 60^{\circ} \quad \text{or} \quad \theta_2 = 30^{\circ}$ Since the angles must be different, we take $\theta_2 = 30^{\circ}$.

Step 3: Calculate the maximum heights for both projectiles
Using the formula for maximum height: $H = \frac{u^2 \sin^2(\theta)}{2g}$ For $\theta_1 = 60^{\circ}$: $H_1 = \frac{(40)^2 \sin^2(60^{\circ})}{2 \times 10} = \frac{1600 \times (\frac{\sqrt{3}}{2})^2}{20} = \frac{1600 \times \frac{3}{4}}{20} = \frac{1200}{20} = 60 \mathrm{~m}$ For $\theta_2 = 30^{\circ}$: $H_2 = \frac{(40)^2 \sin^2(30^{\circ})}{2 \times 10} = \frac{1600 \times (\frac{1}{2})^2}{20} = \frac{1600 \times \frac{1}{4}}{20} = \frac{400}{20} = 20 \mathrm{~m}$

Step 4: Compute the sum of the maximum heights
$H_1 + H_2 = 60 \mathrm{~m} + 20 \mathrm{~m} = 80 \mathrm{~m}$

1. Misidentifying the second angle: Students often forget that $\sin(2\theta)$ is symmetric about $90^{\circ}$, leading to two possible angles for the same range. They might incorrectly assume the second angle is $90^{\circ} - 60^{\circ} = 30^{\circ}$ without verifying the range formula, or they might consider only one solution to $\sin(2\theta_1) = \sin(2\theta_2)$.

2. Incorrect use of the maximum height formula: Some students confuse the formula for maximum height with the formula for time of flight or range. Remember that the maximum height depends on $\sin^2(\theta)$, not $\sin(2\theta)$.

3. Arithmetic errors: Calculating $\sin^2(60^{\circ})$ or $\sin^2(30^{\circ})$ can lead to mistakes if not done carefully. Always double-check the trigonometric values: - $\sin(30^{\circ}) = \frac{1}{2}$ - $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Exam Tip: When two projectiles have the same range with the same initial speed, their angles of projection are complementary. This is a quick way to find the second angle without solving the range equation explicitly.