JEE PYQ: Motion in a Plane - Question ID 219add5c5413 (JEE Main 2022)

Core Formula & Concept:

In this problem, we analyze the relative motion of two objects—the fighter jet and the bullet—moving in a plane. The key concept is relative velocity and the independence of horizontal and vertical motions in projectile motion.

The bullet is fired at an angle $\theta$ with an initial speed $v_b = 400 \text{ m/s}$. The jet is moving horizontally at $v_j = 200 \text{ m/s}$. For the bullet to hit the jet, their relative position must coincide at some time $t$ after firing.

We use the following kinematic equations for projectile motion:

• Horizontal displacement of bullet: $x_b = v_b \cos \theta \cdot t$
• Vertical displacement of bullet: $y_b = v_b \sin \theta \cdot t - \frac{1}{2} g t^2$
• Horizontal displacement of jet: $x_j = v_j \cdot t$ (since it's directly overhead at $t=0$)
• Vertical displacement of jet: $y_j = 0$ (jet flies horizontally at constant altitude)

For the bullet to hit the jet, two conditions must be satisfied:

1. Horizontal coincidence: $x_b = x_j$
2. Vertical coincidence: $y_b = y_j = 0$ (since the jet remains at the same altitude)

From these, we derive the necessary angle $\theta$.

--- Step-by-Step Derivation:

Step 1: Horizontal Coincidence
The bullet and jet must meet horizontally at time $t$: $x_b = x_j \Rightarrow v_b \cos \theta \cdot t = v_j \cdot t$ Assuming $t \neq 0$, we can divide both sides by $t$: $v_b \cos \theta = v_j$ Substitute $v_b = 400 \text{ m/s}$ and $v_j = 200 \text{ m/s}$: $400 \cos \theta = 200 \Rightarrow \cos \theta = \frac{200}{400} = \frac{1}{2}$ This gives: \theta = 60^\circ \quad \text{(since\cos 60^\circ = \frac{1}{2})}

Step 2: Vertical Coincidence (Verification)
We must ensure that at the same time $t$, the bullet returns to the same altitude (i.e., $y_b = 0$). Using the vertical motion equation: $y_b = v_b \sin \theta \cdot t - \frac{1}{2} g t^2 = 0$ Factor out $t$: $t \left( v_b \sin \theta - \frac{1}{2} g t \right) = 0$ Solutions are $t = 0$ (initial time) or: $t = \frac{2 v_b \sin \theta}{g}$ From Step 1, we have $t = \frac{x_j}{v_j} = \frac{v_b \cos \theta \cdot t}{v_j}$. However, we already used horizontal coincidence to find $\theta$. Now, substitute $\theta = 60^\circ$ to verify consistency: $t = \frac{2 \cdot 400 \cdot \sin 60^\circ}{9.8} = \frac{800 \cdot \frac{\sqrt{3}}{2}}{9.8} = \frac{400 \sqrt{3}}{9.8} \approx 70.7 \text{ s}$ At this time, the horizontal distance covered by the jet is: $x_j = v_j \cdot t = 200 \cdot 70.7 \approx 14140 \text{ m}$ The bullet's horizontal distance is: $x_b = v_b \cos \theta \cdot t = 400 \cdot \frac{1}{2} \cdot 70.7 = 14140 \text{ m}$ Both distances match, confirming that $\theta = 60^\circ$ satisfies both horizontal and vertical coincidence.

Conclusion:
The angle $\theta$ at which the bullet must be fired is $\boxed{60^\circ}$.

--- Common Traps & Exam Tip:

Trap 1: Ignoring Relative Motion
Students often forget that the jet is moving and treat the problem as a stationary target. This leads to incorrect assumptions about the bullet's trajectory. Always consider the relative velocity between the projectile and the target.

Trap 2: Overcomplicating Vertical Motion
Some students unnecessarily solve for time $t$ using vertical motion, even though horizontal coincidence alone suffices to find $\theta$. While verification is good, the primary solution comes from horizontal motion.

Trap 3: Misapplying Trigonometry
Students may confuse $\sin \theta$ and $\cos \theta$ when resolving velocity components. Remember: $\cos \theta$ is for horizontal motion, and $\sin \theta$ is for vertical motion.

Exam Tip:
For projectile-hitting-moving-target problems, always:

1. Write down the position equations for both objects.
2. Set their positions equal at time $t$.
3. Solve for the unknown (here, $\theta$) using horizontal motion first, then verify with vertical motion.