JEE PYQ: Motion in a Straight Line - Question ID 0c2d859eb175 (JEE Main 2022)

Core Formula & Concept:

When a ball is dropped from rest under gravity, it undergoes uniformly accelerated motion in a straight line. The key formulas governing this motion are:

• Velocity as a function of time: $v = u + at$, where $u$ is initial velocity, $a$ is acceleration, and $t$ is time.
• Velocity as a function of displacement: $v^2 = u^2 + 2as$, where $s$ is displacement.
• Displacement as a function of time: $s = ut + \frac{1}{2}at^2$.

Here, the ball is dropped from height $h_0 = 10$ m, so initial velocity $u = 0$. The acceleration due to gravity is $g = 10$ m/s² downward. We are asked to find the height $h$ at which the magnitude of velocity equals the magnitude of acceleration due to gravity, i.e., $v = g = 10$ m/s.

Step-by-Step Derivation:

Step 1: Define Variables and Setup

Let $h$ be the height above the ground at which the velocity $v$ equals $g = 10$ m/s. Since the ball is falling downward, the displacement from the initial drop point to this height is $s = h_0 - h = 10 - h$.

Step 2: Apply the Velocity-Displacement Relation

Using the kinematic equation for velocity in terms of displacement: $v^2 = u^2 + 2as$ Since $u = 0$, $a = g = 10$ m/s², and $s = 10 - h$, we substitute: $v^2 = 0 + 2g(10 - h)$ We are given that $v = g = 10$ m/s, so: $(10)^2 = 2 \cdot 10 \cdot (10 - h)$

Step 3: Solve for $h$

Simplify the equation: $100 = 20(10 - h)$ Divide both sides by 20: $5 = 10 - h$ Rearrange: $h = 10 - 5 = 5 \text{ m}$

Step 4: Verification

At $h = 5$ m, the ball has fallen $5$ m from the initial height. Using $v^2 = 2gs$: $v^2 = 2 \cdot 10 \cdot 5 = 100 \implies v = 10 \text{ m/s}$ This matches the condition $v = g$, confirming our solution.

Common Traps & Exam Tip:

Trap 1: Misinterpreting the condition. Students often confuse the condition $v = g$ with $v = u + gt$. While $v = gt$ gives the velocity at time $t$, the question asks for the height where $v = g$, not the time. Using $v = gt$ leads to incorrect results because it doesn't directly relate velocity to height.

Trap 2: Incorrect displacement sign. The displacement $s$ is the distance fallen, so $s = h_0 - h$. Some students mistakenly use $s = h$, leading to wrong calculations.

Exam Tip: Always use the velocity-displacement relation $v^2 = u^2 + 2as$ for problems involving velocity and height. It avoids the need to calculate time and simplifies the solution.