JEE PYQ: Motion in a Plane - Question ID ace538c6e921 (JEE Main 2024)

Core Formula & Concept:

In projectile motion launched horizontally from a height, the motion can be resolved into two independent components:

• Horizontal motion: Uniform motion with constant velocity $v_x$, since there is no acceleration in the horizontal direction (ignoring air resistance).
• Vertical motion: Free-fall under gravity with initial vertical velocity $v_y = 0$. The vertical displacement $H$ is governed by the kinematic equation: $H = \frac{1}{2} g t^2$ where $g$ is the acceleration due to gravity, and $t$ is the time of flight.

The horizontal range $R$ (distance from the foot of the tower where the body lands) is given by: $R = v_x \cdot t$ Since $v_x = v$ (the initial horizontal velocity), and $t$ is the time to fall through height $H$, we can express $t$ from the vertical motion and substitute into the range formula.

Step-by-Step Derivation:

Step 1: Analyze the first scenario (mass $M$, height $H$, velocity $v$, range $100 \text{ m}$)

From the vertical motion: $H = \frac{1}{2} g t_1^2 \implies t_1 = \sqrt{\frac{2H}{g}}$ From the horizontal motion: $R_1 = v \cdot t_1 = v \sqrt{\frac{2H}{g}} = 100 \text{ m}$ So, $v \sqrt{\frac{2H}{g}} = 100 \quad \text{(Equation 1)}$

Step 2: Analyze the second scenario (mass $2M$, height $4H$, velocity $\frac{v}{2}$)

Note: Mass does not affect the trajectory in projectile motion (ignoring air resistance), so $2M$ is irrelevant to the calculation.

From the vertical motion: $4H = \frac{1}{2} g t_2^2 \implies t_2 = \sqrt{\frac{2 \cdot 4H}{g}} = \sqrt{\frac{8H}{g}} = 2 \sqrt{\frac{2H}{g}} = 2 t_1$

From the horizontal motion: $R_2 = \frac{v}{2} \cdot t_2 = \frac{v}{2} \cdot 2 t_1 = v t_1$ But from Equation 1, $v t_1 = 100 \text{ m}$, so: $R_2 = 100 \text{ m}$

Conclusion: The body will touch the ground at a distance of 100 m.

Common Traps & Exam Tip:

• Ignoring mass independence: Many students mistakenly think that mass affects the range. In projectile motion under gravity (no air resistance), mass cancels out and does not influence the trajectory.
• Incorrect time calculation: Students may confuse the time of flight for different heights. Remember: time of flight depends only on the vertical motion and is proportional to the square root of the height.
• Algebraic oversight: Failing to recognize that $\sqrt{4H} = 2\sqrt{H}$, leading to incorrect simplification. Always simplify radicals carefully.

Exam Tip: In such problems, always write down the knowns and unknowns, separate horizontal and vertical motions, and use the time of flight as the bridge between them. Mass is a red herring here—focus on velocity and height.