JEE PYQ: Motion in a Straight Line - Question ID 077d8a0dda5d (JEE Main 2022)

Core Formula & Concept:

This problem combines two fundamental concepts in kinematics:

1. Free-fall under gravity: When the mango is released, it falls vertically downward with constant acceleration $g = 9.8 \, \mathrm{m/s^2}$. The vertical displacement $y$ of the mango in time $t$ is given by: $y = \frac{1}{2} g t^2$ Here, $y = 19.6 \, \mathrm{m}$, so we can solve for $t$, the time the mango takes to reach the ground. 2. Uniform horizontal motion: The cadet is moving horizontally at a constant speed $v = 9 \, \mathrm{km/h}$. We convert this to $\mathrm{m/s}$: $v = 9 \times \frac{1000}{3600} = 2.5 \, \mathrm{m/s}$ The horizontal distance $x$ covered by the cadet in time $t$ is: $x = v \cdot t$ This $x$ is the distance of the cadet from the tree at the instant the mango is dropped, such that the mango lands exactly on the cadet.

The key insight is that the mango and the cadet must meet at the same point at the same time. The time $t$ is determined solely by the vertical fall, and the horizontal distance $x$ is determined by the cadet’s speed over that same time.

Step-by-Step Derivation:

Step 1: Convert the cadet’s speed to SI units

$v = 9 \, \mathrm{km/h} = 9 \times \frac{1000}{3600} = 2.5 \, \mathrm{m/s}$

Step 2: Calculate the time $t$ for the mango to fall 19.6 m

Using the free-fall equation: $y = \frac{1}{2} g t^2$ Substitute $y = 19.6 \, \mathrm{m}$ and $g = 9.8 \, \mathrm{m/s^2}$: $$19.6 = \frac{1}{2} \times 9.8 \times t^2 \\ 19.6 = 4.9 t^2 \\ t^2 = \frac{19.6}{4.9} = 4 \\ t = 2 \, \mathrm{s}$$

Step 3: Calculate the horizontal distance $x$ covered by the cadet in 2 s

Using the uniform motion equation: $x = v \cdot t = 2.5 \times 2 = 5 \, \mathrm{m}$

Step 4: Match the result with the given options

The calculated distance is $5 \, \mathrm{m}$, which corresponds to option A. Common Traps & Exam Tip:

1. Unit inconsistency: Many students forget to convert the cadet’s speed from $\mathrm{km/h}$ to $\mathrm{m/s}$. This leads to incorrect distance calculations. Always ensure all units are consistent (preferably SI units) before substituting values into equations. 2. Misinterpreting the scenario: Some students assume the cadet is stationary or that the mango has an initial horizontal velocity. The mango is dropped (not thrown), so its initial horizontal velocity is zero. The cadet’s motion is purely horizontal and uniform. 3. Incorrect use of kinematic equations: Students sometimes use the wrong equation for free-fall, such as $v = u + at$, when displacement is the required quantity. Always identify what is given and what is asked before selecting the appropriate equation. 4. Sign errors in displacement: While the direction of motion doesn’t affect the magnitude of the answer here, it’s good practice to define a coordinate system (e.g., downward as positive) to avoid confusion in more complex problems.

Exam Tip: In problems involving relative motion or projectile motion, always break the problem into horizontal and vertical components. Solve for the time using the vertical motion (since gravity acts vertically), then use that time to find the horizontal distance. This approach simplifies the problem and reduces errors.