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

Options

A. (a) ` pi (v ^ 2 ) /( g ) `
B. (b)` pi (v^4 ) /(g ^ 2 ) `
C. (c)` (pi ) /(2) (v^4 )/(g ^ 4 ) `
D. (d)` pi (v^ 2 ) /(g ^ 2 ) `

Explanation

To solve the problem of determining the total area around a water fountain that gets wet when water is sprayed out at a speed \( v \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A water fountain sprays water at a speed \( v \).
- We need to find the total area around the fountain that gets wet.

2. **Finding the Maximum Range**:
- The water will follow a projectile motion. The maximum horizontal distance (range) \( R \) that the water can reach is given by the formula:
\[ R = \frac{v^2 \sin 2\theta}{g} \]
- For a fountain, the optimal angle \( \theta \) for maximum range is \( 45^\circ \), where \( \sin 2\theta = \sin 90^\circ = 1 \).
- Therefore, the maximum range simplifies to:
\[ R_{\text{max}} = \frac{v^2}{g} \]

3. **Calculating the Area**:
- The area \( A \) that gets wet around the fountain can be modeled as a circle with radius \( R_{\text{max}} \).
- The area of a circle is given by the formula:
\[ A = \pi R^2 \]
- Substituting \( R_{\text{max}} \) into the area formula:
\[ A = \pi \left(\frac{v^2}{g}\right)^2 \]
- Simplifying this expression:
\[ A = \pi \frac{v^4}{g^2} \]

4. **Final Result**:
- The total area around the fountain that gets wet is:
\[ A = \frac{\pi v^4}{g^2} \]

### Conclusion:
The total area around the fountain that gets wet is given by \( A = \frac{\pi v^4}{g^2} \).