JEE PYQ: Motion in a Straight Line - Question ID 3bcc2581db85 (JEE Main 2026)

Core Formula & Concept:

When a stone is dropped from a moving balloon, it inherits the balloon’s instantaneous velocity at the moment of release. The motion of the stone thereafter is governed by:

• Uniformly accelerated motion under gravity (free fall).
• The balloon continues to ascend at constant velocity.

Key formulas:

• Displacement under constant acceleration: $s = ut + \tfrac{1}{2} a t^2$ where $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.
• Velocity under constant acceleration: $v = u + a t$

Here $g = 10\ \mathrm{m/s^2}$ downward, so for upward motion $a = -g$, and for downward motion $a = +g$.

Step-by-Step Derivation:

Step 1: Identify initial conditions at release

• Balloon height at release: $h_0 = 75\ \mathrm{m}$.
• Balloon velocity (upward): $u_b = 10\ \mathrm{m/s}$.
• Stone inherits this velocity: $u_s = +10\ \mathrm{m/s}$ (upward).

Step 2: Write the stone’s displacement equation

Let $t$ be the time from release until the stone hits the ground. The stone’s displacement from the release point is: $\Delta y = u_s t - \tfrac{1}{2} g t^2$ Since it falls to the ground, the total displacement from the ground is: $h_0 + \Delta y = 0 \quad\Longrightarrow\quad 75 + 10 t - 5 t^2 = 0$

Step 3: Solve the quadratic for $t$

Rearrange: $5 t^2 - 10 t - 75 = 0 \quad\Longrightarrow\quad t^2 - 2 t - 15 = 0$ Solutions: $t = \frac{2 \pm \sqrt{4 + 60}}{2} = \frac{2 \pm 8}{2} \quad\Longrightarrow\quad t = 5\ \mathrm{s}\quad(\text{positive root})$

Step 4: Compute balloon’s height at $t = 5\ \mathrm{s}$

Balloon ascends at constant speed $u_b = 10\ \mathrm{m/s}$, so its additional height is: $\Delta h = u_b t = 10 \times 5 = 50\ \mathrm{m}$ Total height of the balloon when the stone hits the ground: $h = h_0 + \Delta h = 75 + 50 = 125\ \mathrm{m}$

Common Traps & Exam Tip:

1. Sign confusion: Students often take $g$ as negative throughout, leading to incorrect quadratic solutions. Always define a consistent sign convention (e.g., upward positive). 2. Ignoring initial velocity: Forgetting that the stone starts with the balloon’s velocity causes underestimation of the time to hit the ground. 3. Balloon’s motion: Some assume the balloon stops when the stone is dropped; it continues upward at constant speed.

Exam Tip: Draw a clear sketch of the motion, label velocities and displacements, and solve the quadratic carefully. The correct answer is 125 m, option D.