JEE PYQ: Motion in a Straight Line - Question ID aa08c2dca467 (JEE Main 2022)

Core Formula & Concept:

To solve this problem, we rely on the kinematic equations of motion under constant acceleration (gravity, $g$, acting downward). The key concepts and formulas are:

• Equation of motion for displacement: $s = ut + \frac{1}{2} a t^2$ Here, $s$ is displacement, $u$ is initial velocity, $a$ is acceleration, and $t$ is time.
• Velocity at maximum height: At the highest point, the velocity becomes zero. Using $v^2 = u^2 + 2as$, with $v = 0$, $a = -g$, and $s = h$, we get: $0 = u^2 - 2gh \Rightarrow u = \sqrt{2gh}$ This gives the initial velocity in terms of maximum height $h$.
• Time to reach height $y$: We want the time taken to reach height $y = \frac{h}{3}$ during ascent and descent. Using the displacement equation: $y = ut - \frac{1}{2} g t^2$ This is a quadratic in $t$, yielding two roots — one for the upward journey (smaller $t$) and one for the downward journey (larger $t$).

Our goal is to find the ratio: $\frac{t_1}{t_2}$ where $t_1$ is the time to reach $\frac{h}{3}$ on the way up, and $t_2$ is the time to reach the same height on the way down.

--- Step-by-Step Derivation:

Let’s denote:

• $u$: initial upward velocity
• $h$: maximum height reached
• $g$: acceleration due to gravity (acting downward)

Step 1: Express $u$ in terms of $h$
At maximum height, velocity $v = 0$. Using $v^2 = u^2 - 2gh$: $0 = u^2 - 2gh \Rightarrow u^2 = 2gh \Rightarrow u = \sqrt{2gh}$

Step 2: Write the displacement equation for height $y = \frac{h}{3}$
We use: $y = ut - \frac{1}{2} g t^2$ Substitute $y = \frac{h}{3}$ and $u = \sqrt{2gh}$: $\frac{h}{3} = \sqrt{2gh} \cdot t - \frac{1}{2} g t^2$ Divide both sides by $h$: $\frac{1}{3} = \sqrt{\frac{2g}{h}} \cdot t - \frac{g}{2h} t^2$ Let’s define a dimensionless time variable $\tau = t \sqrt{\frac{g}{2h}}$. Then: $\frac{g}{2h} t^2 = \left( t \sqrt{\frac{g}{2h}} \right)^2 = \tau^2$ and $\sqrt{\frac{2g}{h}} \cdot t = \sqrt{\frac{2g}{h}} \cdot \frac{\tau}{\sqrt{g/(2h)}} = \sqrt{\frac{2g}{h}} \cdot \tau \cdot \sqrt{\frac{2h}{g}} = 2\tau$ So the equation becomes: $\frac{1}{3} = 2\tau - \tau^2$ Rearranged: $\tau^2 - 2\tau + \frac{1}{3} = 0$

Step 3: Solve the quadratic equation for $\tau$
$\tau^2 - 2\tau + \frac{1}{3} = 0$ Multiply through by 3 to eliminate fraction: $3\tau^2 - 6\tau + 1 = 0$ Use quadratic formula: $\tau = \frac{6 \pm \sqrt{36 - 12}}{6} = \frac{6 \pm \sqrt{24}}{6} = \frac{6 \pm 2\sqrt{6}}{6} = 1 \pm \frac{\sqrt{6}}{3}$ Simplify $\sqrt{24} = 2\sqrt{6}$, so: $\tau = 1 \pm \frac{\sqrt{6}}{3} = \frac{3 \pm \sqrt{6}}{3}$

Step 4: Identify $\tau_1$ and $\tau_2$
The two roots correspond to the two times when the ball is at $\frac{h}{3}$:

• $\tau_1 = \frac{3 - \sqrt{6}}{3}$ (smaller time, on the way up)
• $\tau_2 = \frac{3 + \sqrt{6}}{3}$ (larger time, on the way down)

Step 5: Compute the ratio $\frac{t_1}{t_2}$
Recall $t = \tau \sqrt{\frac{2h}{g}}$, so: $\frac{t_1}{t_2} = \frac{\tau_1}{\tau_2} = \frac{3 - \sqrt{6}}{3 + \sqrt{6}}$ Now rationalize the denominator: Multiply numerator and denominator by $3 - \sqrt{6}$: $\frac{(3 - \sqrt{6})^2}{(3)^2 - (\sqrt{6})^2} = \frac{9 - 6\sqrt{6} + 6}{9 - 6} = \frac{15 - 6\sqrt{6}}{3} = 5 - 2\sqrt{6}$ Wait — this seems messy. Let’s re-express the ratio in a cleaner form.

Instead, let’s write: $\frac{3 - \sqrt{6}}{3 + \sqrt{6}} = \frac{\sqrt{3} \cdot \sqrt{3} - \sqrt{6}}{\sqrt{3} \cdot \sqrt{3} + \sqrt{6}}$ But $\sqrt{6} = \sqrt{2} \cdot \sqrt{3}$, so: $= \frac{\sqrt{3}(\sqrt{3} - \sqrt{2})}{\sqrt{3}(\sqrt{3} + \sqrt{2})} = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$ This matches option B.

Conclusion: The ratio of the times is: $\frac{t_{\text{up}}}{t_{\text{down}}} = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

--- Common Traps & Exam Tip:

Trap 1: Misidentifying the roots.
Students often confuse which root corresponds to upward and downward motion. The smaller root is always the time during ascent, and the larger one during descent.

Trap 2: Forgetting to rationalize or simplify.
The answer appears in a form that must be simplified or rationalized to match the given options. Directly comparing $\frac{3 - \sqrt{6}}{3 + \sqrt{6}}$ with options is difficult — it must be rewritten in terms of $\sqrt{2}$ and $\sqrt{3}$.

Trap 3: Using wrong sign for acceleration.
It’s crucial to take $g$ as negative when upward is positive. A sign error here leads to incorrect quadratic solutions.

Exam Tip:
Always express time in terms of a dimensionless variable (like $\tau$) to simplify algebra. Also, verify units and consistency — the ratio must be dimensionless and less than 1, which all options satisfy.

Final Answer: Option B is correct.