JEE PYQ: Motion in a Plane - Question ID 2ee841d79d26 (JEE Main 2023)

Core Formula & Concept:

This problem involves projectile motion in a plane, specifically a horizontally launched projectile from a height. The key concepts are:

• Independence of horizontal and vertical motions: The horizontal ($x$) and vertical ($y$) components of motion are independent of each other under gravity (neglecting air resistance).
• Horizontal motion: Since no horizontal force acts (air resistance neglected), the horizontal velocity remains constant: $v_x = u_x = \text{initial horizontal speed} = 5\ \mathrm{ms}^{-1}$
• Vertical motion: The stone is subject to constant downward acceleration due to gravity ($g = 10\ \mathrm{ms}^{-2}$). The vertical velocity increases from zero (since the stone is thrown horizontally) as it falls: $v_y = \sqrt{2gh}$ where $h = 10\ \mathrm{m}$ is the height of the cliff.
• Resultant velocity on impact: The speed of the stone when it hits the ground is the magnitude of the velocity vector, combining horizontal and vertical components: $v = \sqrt{v_x^2 + v_y^2}$

Step-by-Step Derivation:

Step 1: Identify given data

• Height of cliff, $h = 10\ \mathrm{m}$
• Initial horizontal speed, $u_x = 5\ \mathrm{ms}^{-1}$
• Acceleration due to gravity, $g = 10\ \mathrm{ms}^{-2}$
• Initial vertical speed, $u_y = 0$ (thrown horizontally)

Step 2: Compute vertical velocity on impact

Using the kinematic equation for free fall from rest: $v_y^2 = u_y^2 + 2gh$ Since $u_y = 0$, $v_y^2 = 0 + 2 \cdot 10 \cdot 10 = 200$ $v_y = \sqrt{200} = 10\sqrt{2} \approx 14.14\ \mathrm{ms}^{-1}$

Step 3: Horizontal velocity remains unchanged

Since no horizontal acceleration occurs, $v_x = u_x = 5\ \mathrm{ms}^{-1}$

Step 4: Compute resultant speed on impact

The speed $v$ is the magnitude of the velocity vector: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{5^2 + (10\sqrt{2})^2} = \sqrt{25 + 200} = \sqrt{225} = 15\ \mathrm{ms}^{-1}$

Step 5: Match with given options

The calculated speed is $15\ \mathrm{ms}^{-1}$, which corresponds to option D.

Common Traps & Exam Tip:

Students often make the following mistakes:

• Ignoring horizontal velocity: Some assume the stone falls straight down and only calculate $v_y$, leading to $\sqrt{200} \approx 14.14$, which is not among the options. This overlooks the horizontal component.
• Incorrectly combining components: Adding $v_x$ and $v_y$ directly ($5 + 14.14 = 19.14$) instead of using the Pythagorean theorem. This is a fundamental error in vector addition.
• Misapplying kinematic equations: Using $v = u + at$ without considering the correct time of flight. While time can be calculated, it’s unnecessary here and adds complexity.

Exam Tip: Always treat horizontal and vertical motions separately, then combine them vectorially at the end. In horizontally launched projectiles, the initial vertical velocity is zero, simplifying calculations.