JEE PYQ: Motion in a Straight Line - Question ID c24bf0c30458 (JEE Main 2022)

Core Formula & Concept:

When a ball is thrown vertically (upward or downward) or simply dropped from a height, its motion is governed by the kinematic equations of uniformly accelerated motion under gravity. The key formulas are:

$s = ut + \frac{1}{2} a t^2$ $v = u + a t$ where:

• $s$ = displacement (positive downward or upward, depending on convention),
• $u$ = initial velocity,
• $a = g$ = acceleration due to gravity (taken as $+9.8\ \text{m/s}^2$ downward),
• $t$ = time taken.

In this problem, we have three scenarios from the same height $h$:

1. Ball thrown upward with speed $u$, reaches ground in $t_1 = 6\ \text{s}$.
2. Ball thrown downward with same speed $u$, reaches ground in $t_2 = 1.5\ \text{s}$.
3. Ball dropped from rest ($u = 0$), reaches ground in $t_3$ (to be found).

We take downward as positive direction for simplicity.

Step-by-Step Derivation:

Step 1: Define Variables and Equations

Let $h$ be the height of the tower, $u$ the initial speed (same for upward and downward throws), and $g = 9.8\ \text{m/s}^2$. For the upward throw: - Initial velocity is $-u$ (since upward is negative in our downward-positive convention). - Displacement $s = +h$ (downward). - Time $t_1 = 6\ \text{s}$. Using $s = ut + \frac{1}{2} g t^2$: $h = (-u)(6) + \frac{1}{2} g (6)^2 \Rightarrow h = -6u + 18g \quad \text{(1)}$ For the downward throw: - Initial velocity is $+u$. - Displacement $s = +h$. - Time $t_2 = 1.5\ \text{s}$. $h = u(1.5) + \frac{1}{2} g (1.5)^2 \Rightarrow h = 1.5u + 1.125g \quad \text{(2)}$

Step 2: Equate Equations (1) and (2)

Since both equal $h$: $-6u + 18g = 1.5u + 1.125g$ Bring all terms to one side: $-6u - 1.5u = 1.125g - 18g \Rightarrow -7.5u = -16.875g \Rightarrow u = \frac{16.875g}{7.5} = 2.25g$ Substitute $g = 9.8\ \text{m/s}^2$: $u = 2.25 \times 9.8 = 22.05\ \text{m/s}$

Step 3: Find Height $h$

Use equation (2): $h = 1.5u + 1.125g = 1.5(22.05) + 1.125(9.8) = 33.075 + 11.025 = 44.1\ \text{m}$

Step 4: Find Time for Ball Dropped from Rest ($u = 0$)

Use $s = ut + \frac{1}{2} g t^2$, with $u = 0$, $s = h = 44.1\ \text{m}$: $44.1 = 0 \cdot t + \frac{1}{2} \times 9.8 \times t^2 \Rightarrow 44.1 = 4.9 t^2 \Rightarrow t^2 = \frac{44.1}{4.9} = 9 \Rightarrow t = \sqrt{9} = 3\ \text{s}$ Common Traps & Exam Tip:

Trap 1: Sign Convention Confusion
Students often mix up the sign of $u$ when throwing upward. If downward is taken as positive, upward velocity must be negative. Incorrect sign leads to wrong $u$ and $h$. Trap 2: Assuming $g$ Value
Some students use $g = 10\ \text{m/s}^2$ for simplicity, which gives $t = 3\ \text{s}$ coincidentally here, but in other problems, it may lead to rounding errors. Always use $g = 9.8\ \text{m/s}^2$ unless specified. Trap 3: Overcomplicating with Two Variables
Some try to solve for $u$ and $h$ separately without realizing that equating the two expressions for $h$ eliminates one variable immediately. Exam Tip:
Always define a consistent sign convention (e.g., downward positive) and stick to it. Write down all given data clearly. Use the two scenarios to eliminate one unknown ($u$ or $h$) and solve systematically.

Final Answer: The third ball released from rest will reach the ground in 3 seconds. ✅