JEE PYQ: Motion in a Plane - Question ID 3859785957f1 (JEE Main 2026)

Core Formula & Concept:

In projectile motion, when a particle is launched from the ground with an initial speed $v$ at an angle $\theta$ with the horizontal, its trajectory is a parabola. The horizontal range $R$ (the farthest horizontal distance the projectile travels before hitting the ground) is given by the formula:

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

Here:

• $v$ is the initial speed of the projectile,
• $\theta$ is the launch angle,
• $g$ is the acceleration due to gravity ($10 \text{ m/s}^2$ in this problem).

The range $R$ depends on $\sin(2\theta)$. The maximum value of $\sin(2\theta)$ is 1, which occurs when $2\theta = 90^\circ$, i.e., $\theta = 45^\circ$. Therefore, the maximum range $R_{\text{max}}$ is achieved when the projectile is fired at $45^\circ$ to the horizontal, and is given by:

$R_{\text{max}} = \frac{v^2}{g}$

This is the key formula we will use to find the initial speed $v$ of the bullets, given that the farthest distance (i.e., maximum range) is 6.4 m.

--- Step-by-Step Derivation:

Given:

• Maximum range (farthest distance bullets reach), $R_{\text{max}} = 6.4 \text{ m}$
• Acceleration due to gravity, $g = 10 \text{ m/s}^2$

Goal: Find the initial speed $v$ of the bullets.

Step 1: Recall the formula for maximum range:

$R_{\text{max}} = \frac{v^2}{g}$

Step 2: Rearrange the formula to solve for $v^2$:

$v^2 = R_{\text{max}} \cdot g$

Step 3: Substitute the given values:

$v^2 = 6.4 \cdot 10 = 64$

Step 4: Take the square root of both sides to find $v$:

$v = \sqrt{64} = 8 \text{ m/s}$

Thus, the speed of the bullets from the gun is $8 \text{ m/s}$.

--- Common Traps & Exam Tip:

1. Misinterpreting the "farthest distance": Some students confuse the maximum range with other quantities like maximum height or time of flight. The farthest distance in this context refers to the maximum horizontal range, which occurs at a $45^\circ$ launch angle.

2. Forgetting the $\sin(2\theta)$ dependence: Students may incorrectly use $R = \frac{v^2 \sin \theta}{g}$ or $R = \frac{v^2 \cos \theta}{g}$, which are not valid formulas for range. The correct formula involves $\sin(2\theta)$.

3. Incorrectly assuming all angles give the same range: The question states that bullets are fired in "all directions," but only the $45^\circ$ launch angle yields the maximum range. Students must recognize that the given 6.4 m is the maximum range, not the range for an arbitrary angle.

4. Arithmetic errors: Simple calculation mistakes (e.g., $6.4 \times 10 = 64$, then $\sqrt{64} = 8$) can lead to incorrect answers. Always double-check arithmetic steps.

Exam Tip: When a projectile is fired with the same speed in all directions, the maximum range is always $\frac{v^2}{g}$. This is a standard result and should be memorized for quick problem-solving in exams.