JEE PYQ: Motion in a Plane - Question ID 303a7a4a080f (JEE Main 2023)

Core Formula & Concept:

In projectile motion, the motion of an object is resolved into two independent components:

• Horizontal motion: Uniform motion with constant velocity $u \cos \theta$ (since no acceleration acts horizontally).
• Vertical motion: Accelerated motion under gravity with initial velocity $u \sin \theta$ and acceleration $-g$.

Key formulas used:

• Velocity at any time: $\vec{v} = (u \cos \theta) \hat{i} + (u \sin \theta - gt) \hat{j}$.
• At the highest point, the vertical component of velocity becomes zero: $v_y = 0$.
• Speed at any point: $v = \sqrt{v_x^2 + v_y^2}$.
• Time of flight: $T = \frac{2 u \sin \theta}{g}$.

Step-by-Step Derivation:

Step 1: Express the velocity at the highest point.
At the highest point of the projectile's trajectory, the vertical component of velocity is zero. Thus, the velocity vector is purely horizontal: $\vec{v}_{\text{highest}} = (u \cos \theta) \hat{i} + 0 \hat{j}$ The speed at this point is given as $\frac{\sqrt{3}}{2} u$. Therefore: $v_{\text{highest}} = u \cos \theta = \frac{\sqrt{3}}{2} u$ Solving for $\cos \theta$: $\cos \theta = \frac{\sqrt{3}}{2}$ Thus, $\theta = 30^\circ$ (since $\cos 30^\circ = \frac{\sqrt{3}}{2}$).

Step 2: Find $\sin \theta$.
Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$: $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

Step 3: Compute the time of flight.
The time of flight $T$ for a projectile is given by: $T = \frac{2 u \sin \theta}{g}$ Substituting $\sin \theta = \frac{1}{2}$: $T = \frac{2 u \cdot \frac{1}{2}}{g} = \frac{u}{g}$

Conclusion:
The time of flight is $\frac{u}{g}$, which corresponds to option A.

Common Traps & Exam Tip:

Trap 1: Misinterpreting the speed at the highest point.
Students often assume that the speed at the highest point is zero, which is incorrect. Only the vertical component of velocity is zero; the horizontal component remains $u \cos \theta$. Always resolve the velocity into components.

Trap 2: Incorrectly calculating $\sin \theta$.
Some students forget to use the Pythagorean identity and instead guess $\sin \theta$ based on $\cos \theta$. Always derive $\sin \theta$ systematically to avoid errors.

Trap 3: Confusing time of flight with time to reach the highest point.
The time of flight is twice the time taken to reach the highest point. Students sometimes use $T = \frac{u \sin \theta}{g}$ (which is the time to reach the highest point) instead of $T = \frac{2 u \sin \theta}{g}$.

Exam Tip:
When given the speed at the highest point, always:

1. Recognize that the vertical velocity is zero at the highest point.
2. Use the horizontal velocity to find $\cos \theta$.
3. Derive $\sin \theta$ using trigonometric identities.
4. Apply the time of flight formula correctly.