A large number of bullets are fired in all directions with the 'same speed `v_(@)`. The maximum area on the ground on which these bullets will spread is

Options

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

Explanation

To solve the problem of finding the maximum area on the ground where bullets fired in all directions with the same speed \( v_0 \) will spread, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Bullets are fired in all directions with the same speed \( v_0 \). The goal is to determine the maximum area on the ground that these bullets can cover.

2. **Modeling the Trajectory**:
- When a bullet is fired at an angle \( \theta \) with respect to the horizontal, it will follow a projectile motion. The range \( R \) of a projectile is given by the formula:
\[ R = \frac{v_0^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity.

3. **Maximizing the Range**:
- To find the maximum range, we need to maximize \( \sin(2\theta) \). The maximum value of \( \sin(2\theta) \) is 1, which occurs when \( 2\theta = 90^\circ \) or \( \theta = 45^\circ \).
- Therefore, the maximum range \( R_{max} \) when \( \theta = 45^\circ \) is:
\[ R_{max} = \frac{v_0^2}{g} \]

4. **Determining the Area**:
- Since the bullets are fired in all directions, the area covered on the ground can be modeled as a circle with radius equal to the maximum range \( R_{max} \).
- The area \( A \) of a circle is given by the formula:
\[ A = \pi R^2 \]
- Substituting \( R_{max} \) into the area formula:
\[ A = \pi \left(\frac{v_0^2}{g}\right)^2 \]
- Simplifying this expression gives:
\[ A = \pi \frac{v_0^4}{g^2} \]

5. **Final Result**:
- Therefore, the maximum area on the ground on which these bullets will spread is:
\[ A = \frac{\pi v_0^4}{g^2} \]