If a projectile is fired at an angle `theta` with the vertical with velocity u, then maximum height attained is given by:-

Options

A. `(u^(2)cos theta)/(2g)`
B. `(u^(2)sin^(2)theta)/(2g)`
C. `(u^(2)sin^(@)theta)/(g)`
D. `(u^(2)cos^(2)theta)/(2g)`

Explanation

To solve the problem of finding the maximum height attained by a projectile fired at an angle \( \theta \) with the vertical with an initial velocity \( u \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Angles**:
- When a projectile is fired at an angle \( \theta \) with the vertical, the angle with the horizontal (x-axis) will be \( 90^\circ - \theta \) or \( \phi = 90^\circ - \theta \).

2. **Components of Initial Velocity**:
- The initial velocity \( u \) can be resolved into vertical and horizontal components:
- Vertical component: \( u_y = u \cos \theta \)
- Horizontal component: \( u_x = u \sin \theta \)

3. **Using the Formula for Maximum Height**:
- The formula for the maximum height \( h \) attained by a projectile is given by:
\[ h = \frac{u_y^2}{2g} \]
- Here, \( g \) is the acceleration due to gravity.

4. **Substituting the Vertical Component**:
- Substitute the vertical component \( u_y \) into the maximum height formula:
\[ h = \frac{(u \cos \theta)^2}{2g} \]

5. **Simplifying the Expression**:
- Simplifying the equation gives:
\[ h = \frac{u^2 \cos^2 \theta}{2g} \]

6. **Final Result**:
- Therefore, the maximum height attained by the projectile is:
\[ h_{max} = \frac{u^2 \cos^2 \theta}{2g} \]

### Conclusion:
The maximum height attained by a projectile fired at an angle \( \theta \) with the vertical with an initial velocity \( u \) is given by:
\[ h_{max} = \frac{u^2 \cos^2 \theta}{2g} \]