JEE PYQ: Motion in a Plane - Question ID 2193aa4e7b8d (JEE Main 2023)

Core Formula & Concept:

In projectile motion, the vertical component of the motion is governed by the equations of uniformly accelerated motion under gravity. The key idea here is that the projectile reaches the same height at two different times during its flight. This symmetry implies that the two times are equidistant from the time at which the projectile reaches its maximum height.

Let:

• $u$ = initial speed of projection (unknown)
• $\theta = 30^\circ$ = angle of projection with the horizontal
• $g = 10 \text{ m s}^{-2}$ = acceleration due to gravity
• $t_1 = 3 \text{ s}$ and $t_2 = 5 \text{ s}$ = two times at which the projectile is at the same height

The vertical displacement $y(t)$ at any time $t$ is given by: $y(t) = u \sin \theta \cdot t - \frac{1}{2} g t^2$ Since $y(t_1) = y(t_2)$, we have: $u \sin \theta \cdot t_1 - \frac{1}{2} g t_1^2 = u \sin \theta \cdot t_2 - \frac{1}{2} g t_2^2$ This equation can be simplified to find a relationship involving $u$.

Step-by-Step Derivation:

Step 1: Equate vertical displacements at $t_1$ and $t_2$
Given $y(t_1) = y(t_2)$, we write: $u \sin 30^\circ \cdot 3 - \frac{1}{2} \cdot 10 \cdot 3^2 = u \sin 30^\circ \cdot 5 - \frac{1}{2} \cdot 10 \cdot 5^2$

Step 2: Simplify the equation
Substitute $\sin 30^\circ = \frac{1}{2}$ and compute the terms: $u \cdot \frac{1}{2} \cdot 3 - 5 \cdot 9 = u \cdot \frac{1}{2} \cdot 5 - 5 \cdot 25$ $\frac{3u}{2} - 45 = \frac{5u}{2} - 125$

Step 3: Solve for $u$
Bring all terms involving $u$ to one side and constants to the other: $\frac{3u}{2} - \frac{5u}{2} = -125 + 45$ $-\frac{2u}{2} = -80$ $-u = -80 \implies u = 80 \text{ m s}^{-1}$

Verification:
The time at which the projectile reaches its maximum height is the average of $t_1$ and $t_2$: $t_{\text{max}} = \frac{t_1 + t_2}{2} = \frac{3 + 5}{2} = 4 \text{ s}$ Using the vertical motion equation at $t = t_{\text{max}}$, the vertical component of velocity becomes zero: $u \sin 30^\circ - g t_{\text{max}} = 0$ $u \cdot \frac{1}{2} - 10 \cdot 4 = 0$ $\frac{u}{2} = 40 \implies u = 80 \text{ m s}^{-1}$ This confirms our result.

Common Traps & Exam Tip:

• Misapplying symmetry: Students often forget that the two times at which the projectile is at the same height are symmetric about the time of maximum height. This symmetry is crucial for solving the problem efficiently.
• Incorrect sign convention: Ensure that the acceleration due to gravity is taken as negative if upward direction is positive (or vice versa). Consistency in sign convention is key.
• Overcomplicating the problem: Avoid unnecessary calculations involving horizontal motion or range. The problem only requires vertical motion analysis.
• Arithmetic errors: Simple arithmetic mistakes in solving the equation for $u$ can lead to incorrect answers. Double-check calculations.

Final Answer: The speed of projection is $80 \text{ m s}^{-1}$.