JEE PYQ: Motion in a Plane - Question ID 24ab50be2d2c (JEE Main 2022)

Core Formula & Concept:

In projectile motion, the time taken to reach the maximum height depends on the vertical component of the initial velocity. The key formulas involved are:

• Time to reach maximum height ($t$): $t = \frac{u \sin \theta}{g}$ where $u$ is the initial velocity, $\theta$ is the angle of projection with the horizontal, and $g$ is the acceleration due to gravity.
• Maximum height ($H$): $H = \frac{u^2 \sin^2 \theta}{2g}$ (Though not directly used here, it is useful for understanding the motion.)

The problem states that two projectiles reach their maximum heights in the same time, despite being launched at different angles ($30^\circ$ and $45^\circ$). This implies that their vertical components of velocity must adjust such that the time to reach the peak is identical.

--- Step-by-Step Derivation:

Let the initial velocities of the two projectiles be $u_1$ (for $30^\circ$) and $u_2$ (for $45^\circ$).

1. Write the time to reach maximum height for both projectiles:
   For the first projectile (angle $30^\circ$): $t_1 = \frac{u_1 \sin 30^\circ}{g}$
   For the second projectile (angle $45^\circ$): $t_2 = \frac{u_2 \sin 45^\circ}{g}$
2. Set the times equal (since they reach maximum height in the same time): $t_1 = t_2$ $\frac{u_1 \sin 30^\circ}{g} = \frac{u_2 \sin 45^\circ}{g}$
   The $g$ cancels out: $u_1 \sin 30^\circ = u_2 \sin 45^\circ$
3. Substitute the values of $\sin 30^\circ$ and $\sin 45^\circ$: $\sin 30^\circ = \frac{1}{2}, \quad \sin 45^\circ = \frac{1}{\sqrt{2}}$
   Substituting: $u_1 \cdot \frac{1}{2} = u_2 \cdot \frac{1}{\sqrt{2}}$
4. Solve for the ratio $u_1 : u_2$: $\frac{u_1}{u_2} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$
   Thus, the ratio of initial velocities is: $u_1 : u_2 = \sqrt{2} : 1$

--- Common Traps & Exam Tip:

Trap 1: Students often confuse the time to reach maximum height with the total time of flight. The total time of flight depends on $2u \sin \theta / g$, but the time to reach maximum height is only $u \sin \theta / g$.

Trap 2: Some students mistakenly assume that the initial velocities must be equal if the times are equal, ignoring the role of the angle. The angle directly affects the vertical component, so the initial velocities must compensate for the difference in $\sin \theta$.

Exam Tip: Always write down the relevant formula first and then substitute the given values. This avoids confusion and ensures that the correct trigonometric values are used. In this case, remembering that $\sin 30^\circ = 1/2$ and $\sin 45^\circ = 1/\sqrt{2}$ is crucial.