JEE PYQ: Motion in a Straight Line - Question ID 48349f2151f9 (JEE Main 2022)

Core Formula & Concept:

When objects move under constant acceleration (here, gravity $g = 10\ \text{ms}^{-2}$), their position and velocity vary with time according to the kinematic equations for uniformly accelerated motion. The key formulas are:

• Displacement under constant acceleration: $s = ut + \tfrac{1}{2} a t^2$ where $s$ is the displacement, $u$ the initial velocity, $a$ the acceleration, and $t$ the time.
• For free fall or vertical motion, $a = g$ (downward positive).
• If an object is released (not thrown), its initial velocity $u = 0$.

In this problem, two balls start from the same height but at different times. We must relate their positions at the instant they meet.

Step-by-Step Derivation:

Step 1: Define variables and coordinate system

• Let the top of the tower be the origin ($y = 0$), and downward be the positive $y$-direction.
• Height of tower: $H = 180\ \text{m}$.
• Meeting point is $100\ \text{m}$ above the ground, so its $y$-coordinate is $y_{\text{meet}} = H - 100 = 180 - 100 = 80\ \text{m}.$
• Ball A is released at $t = 0$ with $u_A = 0$.
• Ball B is thrown downward at $t = 2\ \text{s}$ with initial velocity $u$.

Step 2: Write position equations

For any time $t \geq 0$:

• Ball A’s position at time $t$: $y_A(t) = \tfrac{1}{2} g t^2 = 5 t^2.$
• Ball B’s motion starts at $t = 2$, so its clock starts at $t' = t - 2$. Its position at time $t \geq 2$ is: $y_B(t) = u (t - 2) + \tfrac{1}{2} g (t - 2)^2 = u (t - 2) + 5 (t - 2)^2.$

Step 3: Set meeting condition

At the meeting instant $t = T$, both balls are at $y = 80\ \text{m}$: $y_A(T) = y_B(T) = 80.$ Thus: $5 T^2 = u (T - 2) + 5 (T - 2)^2. \quad (1)$

Step 4: Solve for $T$

Expand the right side: $5 T^2 = u (T - 2) + 5 (T^2 - 4 T + 4).$ $5 T^2 = u T - 2 u + 5 T^2 - 20 T + 20.$ Cancel $5 T^2$ from both sides: $0 = u T - 2 u - 20 T + 20.$ Rearrange: $u T - 2 u - 20 T + 20 = 0.$ $u (T - 2) - 20 (T - 1) = 0.$ Solve for $u$: $u (T - 2) = 20 (T - 1).$ $u = \frac{20 (T - 1)}{T - 2}. \quad (2)$

Step 5: Use Ball A’s position to find $T$

Ball A reaches $y = 80$ at $T$: $5 T^2 = 80 \Rightarrow T^2 = 16 \Rightarrow T = 4\ \text{s}.$

Step 6: Substitute $T = 4$ into equation (2)

$u = \frac{20 (4 - 1)}{4 - 2} = \frac{20 \times 3}{2} = 30\ \text{ms}^{-1}.$ Common Traps & Exam Tip:

Students often confuse the time origins for the two balls. Ball B starts at $t = 2$, so its motion equation must use $t - 2$, not $t$. Another frequent error is misinterpreting the meeting height: the problem states “100 m above the ground,” which translates to $80\ \text{m}$ below the top, not $100\ \text{m}$ below the top. Always double-check the coordinate system and time offsets.