JEE PYQ: Motion in a Plane - Question ID 3ee74dce3c39 (JEE Main 2019)

This problem involves the concept of projectile motion in a plane, specifically the range of a projectile fired from ground level. The key physics principles and formulas are:

• Range of a Projectile: The horizontal distance traveled by a projectile launched with speed $v$ at an angle $\theta$ with the horizontal (ignoring air resistance) is given by: $R = \frac{v^2 \sin(2\theta)}{g}$ where $g$ is the acceleration due to gravity.
• Maximum Range: The range $R$ is maximized when $\sin(2\theta) = 1$, i.e., when $\theta = 45^\circ$. Thus, the maximum range is: $R_{\text{max}} = \frac{v^2}{g}$
• Area Covered on the Ground: When bullets are fired in all possible directions from a point on horizontal ground, the set of all landing points forms a circular region on the ground. The radius of this circle is the maximum range $R_{\text{max}}$. Therefore, the area covered is: $A = \pi R_{\text{max}}^2 = \pi \left( \frac{v^2}{g} \right)^2 = \pi \frac{v^4}{g^2}$

Thus, the area covered is proportional to the fourth power of the initial speed $v$.

Let’s denote:

• $v_A = 1 \text{ km/s}$: speed of bullets from gun A
• $v_B = 2 \text{ km/s}$: speed of bullets from gun B
• $g$: acceleration due to gravity (same for both)

Step 1: Find maximum range for each gun

For gun A: $R_{A,\text{max}} = \frac{v_A^2}{g} = \frac{(1)^2}{g} = \frac{1}{g}$

For gun B: $R_{B,\text{max}} = \frac{v_B^2}{g} = \frac{(2)^2}{g} = \frac{4}{g}$

Step 2: Find area covered on the ground

The bullets fired in all directions land within a circle of radius $R_{\text{max}}$. So, the area covered is:

For gun A: $A_A = \pi R_{A,\text{max}}^2 = \pi \left( \frac{1}{g} \right)^2 = \pi \frac{1}{g^2}$

For gun B: $A_B = \pi R_{B,\text{max}}^2 = \pi \left( \frac{4}{g} \right)^2 = \pi \frac{16}{g^2}$

Step 3: Compute the ratio of areas

Thus, the ratio of maximum areas covered is $1 : 16$.

Conclusion: The correct answer is A: 1 : 16.

Trap 1: Confusing range with area
Many students compute the maximum range and take the ratio $R_A : R_B = 1 : 4$, then mistakenly assume the area ratio is the same. But area depends on $R^2$, so the ratio becomes $1 : 16$. Students often forget to square the range when computing area.

Trap 2: Ignoring the "all possible directions" condition
Some students think only about one direction (e.g., 45°) and miss that the question asks for the area covered by bullets fired in all directions. This leads them to consider only linear distances, not circular coverage.

Trap 3: Misapplying the range formula
Students sometimes use $R = \frac{v^2 \sin \theta}{g}$ instead of $\sin(2\theta)$, leading to incorrect range calculations.

Exam Tip:
Always remember: When projectiles are fired in all directions from a point on level ground, the landing points form a circle with radius equal to the maximum range. The area of this circle is proportional to $v^4$, not $v^2$. This is a key insight for such problems.