JEE PYQ: Motion in a Straight Line - Question ID 667801bfc50f (JEE Main 2022)

Core Formula & Concept:

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

• Vertical motion under gravity: When an object is thrown upward, its acceleration is $-g$ (assuming upward as positive). The velocity decreases linearly with time until it momentarily becomes zero at the highest point.
• Time to reach maximum height ($t_h$): If a ball is thrown upward with initial velocity $u$, the time to reach the highest point is given by: $v = u - g t_h = 0 \quad \Rightarrow \quad t_h = \frac{u}{g}$
• Maximum height ($H$): Using the displacement equation: $H = u t_h - \frac{1}{2} g t_h^2$ Substituting $t_h = \frac{u}{g}$: $H = u \left(\frac{u}{g}\right) - \frac{1}{2} g \left(\frac{u}{g}\right)^2 = \frac{u^2}{g} - \frac{u^2}{2g} = \frac{u^2}{2g}$ So, $H = \frac{u^2}{2g}$
• Frequency of throwing balls ($n$): The juggler throws $n$ balls per second. This means the time interval between consecutive throws is: $\Delta t = \frac{1}{n}$
• Key insight: The problem states that the next ball is thrown when the first ball reaches its highest point. This implies that the time between throws ($\Delta t$) is equal to the time taken by a ball to reach its maximum height ($t_h$). Therefore: $\Delta t = t_h \quad \Rightarrow \quad \frac{1}{n} = \frac{u}{g}$ From this, we can solve for $u$: $u = \frac{g}{n}$

Step-by-Step Derivation:

We now derive the maximum height $H$ in terms of $g$ and $n$.

1. From the key insight above, we have: $u = \frac{g}{n}$
2. Substitute $u$ into the expression for maximum height: $H = \frac{u^2}{2g} = \frac{\left(\frac{g}{n}\right)^2}{2g} = \frac{g^2 / n^2}{2g} = \frac{g}{2 n^2}$
3. Thus, the maximum height reached by the balls is: $H = \frac{g}{2 n^2}$
4. Comparing with the given options, this matches option D.

Common Traps & Exam Tip:

Students often make the following mistakes:

• Misinterpreting the throwing frequency: Some confuse $n$ (balls per second) with the time interval. They may incorrectly set $t_h = n$ instead of $t_h = \frac{1}{n}$. Always remember: if $n$ balls are thrown per second, the time between throws is $\frac{1}{n}$.
• Incorrect sign convention: Taking $g$ as positive when upward is positive leads to errors in the kinematic equations. Always define a consistent sign convention (e.g., upward positive, so acceleration is $-g$).
• Confusing $u$ and $H$: Some students directly equate $H = \frac{g}{n}$, ignoring the $\frac{u^2}{2g}$ formula. Always derive $H$ from first principles.
• Algebraic errors: Squaring $\frac{g}{n}$ and simplifying $\frac{g^2}{n^2} \div 2g$ is a common source of mistakes. Double-check each algebraic step.

Exam Tip: In juggling or periodic motion problems, always relate the time between events (here, throws) to the kinematic time (here, time to reach max height). This connection is the key to solving such problems.