JEE PYQ: Motion in a Plane - Question ID 99f9709770f8 (JEE Main 2023)

Core Formula & Concept:

In projectile motion, the trajectory of a particle projected from the ground can be expressed as a quadratic equation in $x$ and $y$. The general form of the trajectory equation is: $y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$ where: - $y$ is the vertical displacement, - $x$ is the horizontal displacement, - $\theta$ is the projection angle, - $u$ is the initial velocity, - $g$ is the acceleration due to gravity.

The given trajectory equation is: $y = x - \frac{x^2}{20}$ This resembles the standard trajectory equation, allowing us to compare coefficients and extract key parameters like the initial velocity and projection angle. However, for finding the maximum height, we can directly analyze the trajectory equation without explicitly determining $u$ or $\theta$.

The maximum height occurs when the vertical component of velocity becomes zero. In terms of the trajectory equation, this corresponds to the vertex of the parabola described by $y$ as a function of $x$. For a quadratic equation of the form $y = ax^2 + bx + c$, the vertex (which gives the maximum or minimum point) occurs at $x = -\frac{b}{2a}$.

Step-by-Step Derivation:

Given the trajectory equation: $y = x - \frac{x^2}{20}$ We can rewrite it in standard quadratic form: $y = -\frac{1}{20} x^2 + x$ Here, $a = -\frac{1}{20}$ and $b = 1$.

The $x$-coordinate of the vertex (where maximum height occurs) is: $x = -\frac{b}{2a} = -\frac{1}{2 \times \left(-\frac{1}{20}\right)} = \frac{1}{\frac{1}{10}} = 10 \text{ m}$

Substitute $x = 10$ m back into the trajectory equation to find the maximum height $y_{\text{max}}$: $y_{\text{max}} = 10 - \frac{(10)^2}{20} = 10 - \frac{100}{20} = 10 - 5 = 5 \text{ m}$

Thus, the maximum height attained by the projectile is 5 m.

Common Traps & Exam Tip:

1. Misidentifying the vertex formula: Students often confuse the vertex formula for $x$ with other quadratic properties. Remember, for $y = ax^2 + bx + c$, the vertex occurs at $x = -\frac{b}{2a}$.


2. Incorrectly comparing coefficients: Some students try to match the given equation with the standard trajectory equation to find $u$ and $\theta$. While this is possible, it is unnecessary for finding the maximum height and can lead to calculation errors.


3. Sign errors in the quadratic term: The negative sign in $-\frac{x^2}{20}$ indicates a downward-opening parabola, confirming that the vertex represents the maximum height. Ignoring the sign can lead to incorrect conclusions about the nature of the extremum.


4. Unit confusion: The problem states that $x$ and $y$ are in meters, so the final answer must be in meters. Ensure that no unit conversions are mistakenly applied.

Exam Tip: For trajectory problems, always check if the question can be solved directly using the properties of quadratic equations before diving into kinematic equations. This saves time and reduces the chance of errors.