JEE PYQ: Motion in a Plane - Question ID 4b6d1e6e38e5 (JEE Main 2025)

Here's a detailed solution to the problem, designed to guide you through the concepts and calculations systematically.

Core Formula & Concept:

This problem deals with projectile motion, which is the motion of an object thrown or projected into the air, subject only to the acceleration of gravity. We assume no air resistance. The key parameters for a projectile launched with initial velocity $u$ at an angle $\theta$ with the horizontal are:

1. Maximum Height (H): This is the highest vertical displacement achieved by the projectile from its point of projection. It depends on the vertical component of the initial velocity. $H = \frac{u^2 \sin^2 \theta}{2g}$ where $g$ is the acceleration due to gravity.
2. Time of Flight (T): This is the total time for which the projectile remains in the air, from launch until it hits the ground at the same horizontal level. It also depends on the vertical component of the initial velocity. $T = \frac{2u \sin \theta}{g}$

The problem states that two balls have the same mass and initial velocity, but are projected at different angles. Mass is irrelevant in ideal projectile motion calculations. The constant initial velocity $u$ and acceleration due to gravity $g$ will be common for both balls.

Step-by-Step Derivation:

Let the projection angles for the first and second balls be $\theta_1$ and $\theta_2$, respectively. Let their initial velocity be $u$.

Step 1: Write down the expressions for Maximum Height and Time of Flight for both balls. For the first ball (angle $\theta_1$): $H_1 = \frac{u^2 \sin^2 \theta_1}{2g}$ $T_1 = \frac{2u \sin \theta_1}{g}$ For the second ball (angle $\theta_2$): $H_2 = \frac{u^2 \sin^2 \theta_2}{2g}$ $T_2 = \frac{2u \sin \theta_2}{g}$ Step 2: Use the given condition relating the maximum heights. The problem states that the maximum height reached by the first ball is 8 times higher than that of the second ball. $H_1 = 8 H_2$ Substitute the expressions for $H_1$ and $H_2$: $\frac{u^2 \sin^2 \theta_1}{2g} = 8 \left( \frac{u^2 \sin^2 \theta_2}{2g} \right)$ Notice that the terms $\frac{u^2}{2g}$ are common on both sides and can be cancelled out: $\sin^2 \theta_1 = 8 \sin^2 \theta_2$ Now, take the square root of both sides. Since $\theta_1$ and $\theta_2$ are projection angles, $\sin \theta_1$ and $\sin \theta_2$ must be positive (for angles between $0^\circ$ and $180^\circ$, which are typical for projectile motion). $\sqrt{\sin^2 \theta_1} = \sqrt{8 \sin^2 \theta_2}$ $\sin \theta_1 = \sqrt{8} \sin \theta_2$ We know that $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$. $\sin \theta_1 = 2\sqrt{2} \sin \theta_2 \quad \text{(Equation 1)}$ Step 3: Find the ratio of the total flying times, $T_1 : T_2$. We need to calculate $\frac{T_1}{T_2}$. Substitute the expressions for $T_1$ and $T_2$: $\frac{T_1}{T_2} = \frac{\frac{2u \sin \theta_1}{g}}{\frac{2u \sin \theta_2}{g}}$ Again, the terms $\frac{2u}{g}$ are common in the numerator and denominator and can be cancelled out: $\frac{T_1}{T_2} = \frac{\sin \theta_1}{\sin \theta_2}$ Now, substitute the relationship we found in Equation 1: $\sin \theta_1 = 2\sqrt{2} \sin \theta_2$. $\frac{T_1}{T_2} = \frac{2\sqrt{2} \sin \theta_2}{\sin \theta_2}$ Cancel out $\sin \theta_2$: $\frac{T_1}{T_2} = 2\sqrt{2}$ Therefore, the ratio of $T_1$ and $T_2$ is $2\sqrt{2} : 1$. Comparing this result with the given options, we find that it matches option C.
Common Traps & Exam Tip: Common Traps:

1. Algebraic Error with Square Roots: A frequent mistake is to incorrectly simplify $\sqrt{8}$ as $4$ or $2$, instead of $2\sqrt{2}$. Be careful with your square root calculations.
2. Forgetting the Square: Students might sometimes forget the square in the $\sin^2\theta$ term for maximum height, leading to incorrect relationships.
3. Confusing Formulas: Ensure you use the correct formulas for maximum height and time of flight. Mixing them up or using a formula for horizontal range (which is not relevant here) can lead to errors.

Exam Tip:

For problems involving ratios in projectile motion, always look for proportionality. Notice that: $H \propto \sin^2 \theta$ $T \propto \sin \theta$ From these two proportionalities, we can deduce a direct relationship between $H$ and $T$. Since $T \propto \sin \theta$, then $\sin \theta \propto T$. Substituting this into the $H$ proportionality: $H \propto (T)^2$ or $T \propto \sqrt{H}$ Using this shortcut:

$\frac{T_1}{T_2} = \sqrt{\frac{H_1}{H_2}}$

Given $H_1 = 8 H_2$, we have $\frac{H_1}{H_2} = 8$.

$\frac{T_1}{T_2} = \sqrt{8} = 2\sqrt{2}$

This quick method can save significant time during the exam. However, make sure you understand the full derivation first before relying solely on shortcuts.

The final answer is \boxed{\text{2\sqrt{2} : 1}}.