JEE PYQ: Motion in a Plane - Question ID 3c79b4b9bf67 (JEE Main 2011)

In projectile motion on a horizontal plane, the range $R$ of a particle launched with speed $v$ at an angle $\theta$ to the horizontal is given by: $R = \frac{v^2 \sin 2\theta}{g}.$ The fountain sprays water in all directions (i.e., for every angle $\theta$ from $0$ to $\pi/2$). The area that gets wet is the union of all possible landing points of the water droplets. Since the droplets are sprayed symmetrically in every direction, the wet region is a circular disk whose radius is the maximum range $R_{\max}$.

Step 1: Express the range as a function of launch angle
For a fixed launch speed $v$, the range is $R(\theta) = \frac{v^2 \sin 2\theta}{g}.$

Step 2: Find the maximum range
The sine function reaches its maximum value of $1$ when $2\theta = \pi/2$, i.e.\ $\theta = \pi/4$. Hence $R_{\max} = \frac{v^2}{g}.$ However, this is only the range for one particular direction. In reality, the fountain sprays water in all directions, so the farthest any droplet can land is still $R_{\max}$, but now in every azimuthal direction.

Step 3: Determine the wet area
Because the fountain is symmetric about its vertical axis, the set of all landing points forms a perfect circle of radius $R_{\max}$. The area $A$ of this circle is $A = \pi R_{\max}^2 = \pi \left(\frac{v^2}{g}\right)^2 = \pi \frac{v^4}{g^2}.$

Step 4: Match with the given options
Comparing with the choices, we see that option A is exactly $\pi \frac{v^4}{g^2}$, which matches our result.

1. Confusing maximum range with average range: Some students compute the average range over all angles and then square it, leading to an incorrect factor of $\frac{1}{2}$. 2. Forgetting the square on the range: The area is $\pi R^2$, not $\pi R$. Squaring $v^2/g$ gives $v^4/g^2$. 3. Ignoring symmetry: One must recognize that the wet region is a full circle, not just a single ray of length $R_{\max}$.

Exam Tip: Always sketch the physical situation. Here, drawing the fountain at the center of a circle whose radius is the maximum range immediately suggests the area formula $\pi R^2$.