JEE PYQ: Motion in a Straight Line - Question ID 105f3bca91ae (JEE Main 2024)

Core Formula & Concept:

When a body moves under constant acceleration (here, gravity $g$), its displacement $s$, initial velocity $u$, time $t$, and acceleration $a$ are related by the kinematic equation: $s = ut + \tfrac{1}{2} a t^2.$ In vertical motion near Earth’s surface, we take $a = g$ (downward) and measure displacements upward as positive. If the body is projected upward or downward from a height $h$, the displacement from the launch point to the ground is $-h$ (since the ground is below the launch point).

Step-by-Step Derivation:

1. Define variables and sign convention
Let the tower height be $h$. Take upward as positive. Then the displacement from the top of the tower to the ground is $s = -h$. The acceleration due to gravity is $a = -g$ (since it acts downward).

2. Upward projection
The body is projected upward with speed $u$. The kinematic equation gives -h = u\,t_1 + \tfrac{1}{2}(-g)\,t_1^2 \quad\Longrightarrow\quad h = -u\,t_1 + \tfrac{1}{2}g\,t_1^2. \tag{1}

3. Downward projection
The body is projected downward with the same speed $u$. Now the initial velocity is $-u$. The equation becomes -h = (-u)\,t_2 + \tfrac{1}{2}(-g)\,t_2^2 \quad\Longrightarrow\quad h = u\,t_2 + \tfrac{1}{2}g\,t_2^2. \tag{2}

4. Eliminate $u$ and $h$
Subtract (1) from (2): $0 = u\,(t_1 + t_2) + \tfrac{1}{2}g\,(t_2^2 - t_1^2) = u\,(t_1 + t_2) + \tfrac{1}{2}g\,(t_2 - t_1)(t_2 + t_1).$ Divide by $t_1 + t_2$ (nonzero): 0 = u + \tfrac{1}{2}g\,(t_2 - t_1) \quad\Longrightarrow\quad u = \tfrac{1}{2}g\,(t_1 - t_2). \tag{3}

5. Express $h$ in terms of $t_1$ and $t_2$
Substitute (3) into (1): h = -\bigl[\tfrac{1}{2}g\,(t_1 - t_2)\bigr]\,t_1 + \tfrac{1}{2}g\,t_1^2 = \tfrac{1}{2}g\,t_1 t_2. \tag{4}

6. Free fall (dropped from rest)
When the body is simply dropped, $u = 0$. The kinematic equation is $-h = 0 + \tfrac{1}{2}(-g)\,t^2 \quad\Longrightarrow\quad h = \tfrac{1}{2}g\,t^2.$ Equate this to (4): $\tfrac{1}{2}g\,t^2 = \tfrac{1}{2}g\,t_1 t_2 \quad\Longrightarrow\quad t^2 = t_1 t_2 \quad\Longrightarrow\quad t = \sqrt{t_1 t_2}.$

Common Traps & Exam Tip:

1. Sign errors: Many students forget to take downward as negative or confuse the sign of $g$. Always fix a sign convention at the start. 2. Algebraic slips: When subtracting the two quadratic equations, it is easy to mis-handle the difference of squares. Factor $t_2^2 - t_1^2 = (t_2 - t_1)(t_2 + t_1)$ carefully. 3. Option confusion: The correct answer is the geometric mean $\sqrt{t_1 t_2}$, not the arithmetic mean or difference. Memorize the result as “the dropped time is the geometric mean of the two projection times.”