JEE PYQ: Motion in a Plane - Question ID ea63353c60cd (JEE Main 2004)

In this problem, we analyze the motion of a projectile (the ball) and the motion of a person running with constant velocity. The key concepts involved are:

• Projectile Motion: When a ball is thrown with an initial speed $v_0$ at an angle $\theta$, its motion can be resolved into horizontal and vertical components.
  • Horizontal component of velocity: $v_{0x} = v_0 \cos \theta$ (constant, since no horizontal acceleration)
  • Vertical component of velocity: $v_{0y} = v_0 \sin \theta$ (changes due to gravity, $g$)
  • Time of flight (total time the ball stays in the air): $T = \frac{2 v_0 \sin \theta}{g}$
  • Horizontal range (distance traveled horizontally by the ball): $R = v_0 \cos \theta \cdot T = \frac{v_0^2 \sin 2\theta}{g}$
• Uniform Motion of the Person: The person runs with a constant speed $\frac{v_0}{2}$ in the same horizontal direction as the ball’s projection. The distance covered by the person in time $t$ is: $d_{\text{person}} = \frac{v_0}{2} \cdot t$
• Condition to Catch the Ball: The person will catch the ball if, at some time $t \leq T$, the horizontal distance covered by the ball equals the distance covered by the person. That is: $v_0 \cos \theta \cdot t = \frac{v_0}{2} \cdot t$ However, this naive approach leads to an inconsistency unless we consider the full trajectory and timing.

The crucial insight is that the person must reach the same horizontal position as the ball at the same time the ball returns to the ground (or earlier, but the ball must be catchable). Since the person starts at the same point and runs at constant speed, the only way to catch the ball is if the person reaches the ball’s landing point exactly when the ball lands.

Thus, the condition becomes: $\text{Distance run by person in time } T = \text{Range of the ball}$ That is: $\frac{v_0}{2} \cdot T = R$ Substituting $T$ and $R$: $\frac{v_0}{2} \cdot \frac{2 v_0 \sin \theta}{g} = \frac{v_0^2 \sin 2\theta}{g}$ Simplify and solve for $\theta$.

Let’s derive the condition step-by-step.

1. Express the time of flight $T$: $T = \frac{2 v_0 \sin \theta}{g}$
2. Express the range $R$ of the projectile: $R = v_0 \cos \theta \cdot T = v_0 \cos \theta \cdot \frac{2 v_0 \sin \theta}{g} = \frac{2 v_0^2 \sin \theta \cos \theta}{g} = \frac{v_0^2 \sin 2\theta}{g}$
3. Distance covered by the person in time $T$: $d_{\text{person}} = \frac{v_0}{2} \cdot T = \frac{v_0}{2} \cdot \frac{2 v_0 \sin \theta}{g} = \frac{v_0^2 \sin \theta}{g}$
4. Set the distance run by the person equal to the range of the ball (condition to catch): $d_{\text{person}} = R \Rightarrow \frac{v_0^2 \sin \theta}{g} = \frac{v_0^2 \sin 2\theta}{g}$ Cancel $\frac{v_0^2}{g}$ from both sides: $\sin \theta = \sin 2\theta$
5. Use trigonometric identity for $\sin 2\theta$: $\sin 2\theta = 2 \sin \theta \cos \theta$ So the equation becomes: $\sin \theta = 2 \sin \theta \cos \theta$
6. Solve for $\theta$: Assuming $\sin \theta \neq 0$ (i.e., $\theta \neq 0^\circ$ or $180^\circ$), we can divide both sides by $\sin \theta$: $1 = 2 \cos \theta \Rightarrow \cos \theta = \frac{1}{2}$ Thus: $\theta = 60^\circ$ (Note: $\theta = 300^\circ$ is not physically meaningful in this context.)
7. Verify the solution: For $\theta = 60^\circ$:
   • $\sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos 60^\circ = \frac{1}{2}$
   • $\sin 2\theta = \sin 120^\circ = \frac{\sqrt{3}}{2} = \sin 60^\circ$
   • Thus, $\sin \theta = \sin 2\theta$ holds true.
8. Conclusion: The person will be able to catch the ball if the angle of projection is $60^\circ$.

Students often make the following mistakes:

• Assuming the person catches the ball at any time $t$: Some students set $v_0 \cos \theta = \frac{v_0}{2}$ and conclude $\cos \theta = \frac{1}{2}$, leading to $\theta = 60^\circ$. While this gives the correct answer, it is incorrect reasoning. The person does not catch the ball at the instant of projection but at the time of landing. The correct approach is to equate the distances covered in the time of flight.
• Ignoring the time of flight: Students may forget that the person must run for the entire duration the ball is in the air. The condition must involve the full trajectory, not just initial velocities.
• Misapplying trigonometric identities: Errors in simplifying $\sin 2\theta$ or canceling $\sin \theta$ without considering $\sin \theta = 0$ can lead to incorrect angles.
• Overlooking the physical meaning of $\theta$: Angles like $0^\circ$ or $90^\circ$ are not valid solutions here, even if they satisfy the equation mathematically.

Exam Tip: Always write down the full expressions for time of flight and range before setting up the condition. This ensures you account for the full motion of the projectile and avoid oversimplifying the problem.

Final Answer: Option C (Yes, $60^\circ$) is correct.