JEE PYQ: Motion in a Plane - Question ID fde15d6fce01 (JEE Main 2023)

Core Formula & Concept:

In projectile motion, the motion can be resolved into two independent components:

• Horizontal motion: Uniform motion with constant velocity since no acceleration acts horizontally (ignoring air resistance). The horizontal component of velocity remains unchanged throughout the flight.
• Vertical motion: Uniformly accelerated motion under gravity ($g$). The vertical component of velocity changes linearly with time due to gravitational acceleration.

The initial velocity $\vec{v}_0$ is given at an angle $\theta$ from the horizontal. Its components are:

• Horizontal component: $v_{0x} = v_0 \cos \theta$
• Vertical component: $v_{0y} = v_0 \sin \theta$

At any time $t$, the velocity components are:

• $v_x(t) = v_{0x} = v_0 \cos \theta$ (constant)
• $v_y(t) = v_{0y} - gt = v_0 \sin \theta - gt$ (changes due to gravity)

The magnitude of the velocity at time $t$ is: $v(t) = \sqrt{v_x(t)^2 + v_y(t)^2}$

Step-by-Step Derivation:

Given data:

• Initial velocity, $v_0 = 40 \text{ m/s}$
• Projection angle, $\theta = 30^\circ$
• Time, $t = 2 \text{ s}$
• Acceleration due to gravity, $g = 10 \text{ m/s}^2$

Step 1: Resolve initial velocity into components

Horizontal component: $v_{0x} = v_0 \cos \theta = 40 \cos 30^\circ = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \text{ m/s}$

Vertical component: $v_{0y} = v_0 \sin \theta = 40 \sin 30^\circ = 40 \times \frac{1}{2} = 20 \text{ m/s}$

Step 2: Compute velocity components at $t = 2 \text{ s}$

Horizontal component remains constant: $v_x(2) = v_{0x} = 20\sqrt{3} \text{ m/s}$

Vertical component changes due to gravity: $v_y(2) = v_{0y} - gt = 20 - (10)(2) = 20 - 20 = 0 \text{ m/s}$

Step 3: Compute magnitude of velocity at $t = 2 \text{ s}$

$v(2) = \sqrt{v_x(2)^2 + v_y(2)^2} = \sqrt{(20\sqrt{3})^2 + 0^2} = \sqrt{400 \times 3} = \sqrt{1200} = 20\sqrt{3} \text{ m/s}$

Step 4: Match with given options

The calculated velocity is $20\sqrt{3} \text{ m/s}$, which corresponds to option A.

Common Traps & Exam Tip:

Students often make the following mistakes:

• Ignoring the independence of horizontal and vertical motions: Some students mistakenly apply acceleration to the horizontal component or forget that horizontal velocity remains constant.
• Misapplying the angle: Confusing $\sin$ and $\cos$ while resolving components, especially at standard angles like $30^\circ$.
• Assuming velocity becomes zero at maximum height: While the vertical component becomes zero at the highest point, the horizontal component remains, so the total velocity is not zero. In this case, at $t = 2 \text{ s}$, the projectile is at its peak (since $v_y = 0$), but the velocity is not zero.
• Incorrectly calculating the magnitude: Forgetting to take the square root or squaring the components incorrectly.

Exam Tip: Always resolve the motion into horizontal and vertical components first. Remember that horizontal velocity is constant, and vertical velocity changes due to gravity. Double-check the trigonometric values for standard angles.