JEE PYQ: Motion in a Straight Line - Question ID 2bb86096b495 (JEE Main 2026)

Core Formula & Concept:

When a ball is dropped from rest under gravity, its motion is uniformly accelerated. The key formulas we use are:

• Velocity as a function of time: $v = u + at$, where $u = 0$ (dropped from rest), $a = g = 10 \, \text{m/s}^2$.
• Displacement as a function of time: $s = ut + \frac{1}{2} a t^2$. Since $u = 0$, this simplifies to $s = \frac{1}{2} g t^2$.
• Velocity as a function of displacement: $v^2 = u^2 + 2 a s$. Again, $u = 0$, so $v^2 = 2 g s$.

The question asks for the height above the ground where the magnitude of velocity equals the magnitude of acceleration due to gravity. Since $g = 10 \, \text{m/s}^2$, we seek the height where $v = 10 \, \text{m/s}$.

Step-by-Step Derivation:

Let’s define:

• $H = 18 \, \text{m}$: Initial height from which the ball is dropped.
• $h$: Height above the ground where $v = g = 10 \, \text{m/s}$.
• $s = H - h$: Distance fallen by the ball when it reaches height $h$.

Using the velocity-displacement relation for free fall: $v^2 = 2 g s$ We substitute $v = 10 \, \text{m/s}$ and $s = 18 - h$: $(10)^2 = 2 \cdot 10 \cdot (18 - h)$ $100 = 20 (18 - h)$

Now, solve for $h$: $100 = 360 - 20 h$ $20 h = 360 - 100$ $20 h = 260$ $h = \frac{260}{20} = 13 \, \text{m}$

Thus, the height above the ground where the velocity equals $10 \, \text{m/s}$ is 13 m.

Common Traps & Exam Tip:

Students often make the following mistakes:

1. Misinterpreting the question: Some confuse the height at which velocity equals $g$ with the height where acceleration equals velocity. Remember, acceleration due to gravity is constant ($g = 10 \, \text{m/s}^2$), but velocity increases as the ball falls.
2. Incorrect sign convention: If upward is taken as positive, velocity should be negative (since the ball is falling downward). However, the question asks for magnitude, so signs can be ignored.
3. Using time-based equations unnecessarily: While $v = gt$ is valid, it requires solving for time first, which is an extra step. The velocity-displacement relation ($v^2 = 2gs$) is more direct.
4. Forgetting to subtract from initial height: The displacement $s$ is the distance fallen, not the height above ground. Always define $s = H - h$ clearly.

Exam Tip: For free-fall problems, always write down known quantities ($u, g, H$) and the unknown ($h$). Use the most efficient kinematic equation to avoid unnecessary steps.