A water fountain on the ground sprinkles water all around it . If the speed of water coming out of the fountain is ` v` , the total area around the fountain that gets wet is :

Options

A. `pi/2 v^(4)/g^(2)`
B. `pi v^(2)/g^(2)`
C. `pi v^(2)/g`
D. `pi v^(4)/g^(2)`

Explanation

To solve the problem regarding the area around a water fountain that gets wet when water is sprinkled out at a speed \( v \), we can approach it step by step.

### Step 1: Understand the Geometry of the Water Fountain
When water is ejected from the fountain, it follows a projectile motion. The water will form a parabolic trajectory as it moves upwards and then falls back down due to gravity.

### Step 2: Determine the Time of Flight
The time of flight \( T \) for a projectile launched at an angle \( \theta \) with an initial speed \( v \) can be calculated using the formula:
\[ T = \frac{2v \sin \theta}{g} \]
where \( g \) is the acceleration due to gravity. For our case, if we assume the water is ejected vertically, \( \theta = 90^\circ \), thus:
\[ T = \frac{2v}{g} \]

### Step 3: Calculate the Horizontal Range
The horizontal distance \( R \) (or range) that the water travels can be calculated using the formula:
\[ R = v \cos \theta \cdot T \]
For vertical ejection, \( \cos 90^\circ = 0 \), thus the horizontal range will be zero. However, if we consider that the water spreads out in all directions, we can think of the effective radius of the area that gets wet.

### Step 4: Determine the Area
The area \( A \) that gets wet can be approximated as a circle with radius \( R \). The area of a circle is given by:
\[ A = \pi R^2 \]
Substituting the expression for \( R \) from the time of flight:
\[ A = \pi \left( \frac{v^2}{g} \right) \]

### Final Expression
Thus, the total area around the fountain that gets wet is:
\[ A = \frac{\pi v^2}{g} \]

### Summary
The total area around the fountain that gets wet when water is sprinkled out at a speed \( v \) is given by:
\[ A = \frac{\pi v^2}{g} \]