A ball is projected horizontal from the top of a tower with a velocity `v_(0)`. It will be moving at an angle of `60^(@)` with the horizontal after time.

Options

A. `(v_(0))/(sqrt(3)g)`
B. `(sqrt(3)v_(0))/(g)`
C. `(v_(0))/(g)`
D. `(v_(0))/(g)`

Explanation

To solve the problem of finding the time at which a ball projected horizontally from the top of a tower makes an angle of 60 degrees with the horizontal, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The ball is projected horizontally with an initial velocity \( u_0 \).
- The horizontal velocity \( V_x \) remains constant throughout the motion since there is no horizontal acceleration.

2. **Identify the Angle**:
- The angle \( \theta \) with the horizontal is given as 60 degrees.

3. **Using Trigonometry**:
- At the angle of 60 degrees, we can relate the vertical and horizontal components of the velocity using the tangent function:
\[ \tan(\theta) = \frac{V_y}{V_x} \]
- Here, \( V_y \) is the vertical component of the velocity, and \( V_x \) is the horizontal component, which is equal to \( u_0 \).

4. **Calculate \( V_y \)**:
- For \( \theta = 60^\circ \):
\[ \tan(60^\circ) = \sqrt{3} \]
- Therefore, we have:
\[ \sqrt{3} = \frac{V_y}{u_0} \]
- Rearranging gives:
\[ V_y = \sqrt{3} u_0 \]

5. **Finding the Vertical Velocity**:
- The vertical velocity \( V_y \) can also be expressed using the equation of motion:
\[ V_y = u_{y} + g t \]
- Since the initial vertical velocity \( u_{y} = 0 \) (the ball is projected horizontally), we have:
\[ V_y = g t \]

6. **Setting the Equations Equal**:
- Now we can set the two expressions for \( V_y \) equal to each other:
\[ \sqrt{3} u_0 = g t \]

7. **Solving for Time \( t \)**:
- Rearranging gives:
\[ t = \frac{\sqrt{3} u_0}{g} \]

### Final Answer:
The time at which the ball makes an angle of 60 degrees with the horizontal is:
\[ t = \frac{\sqrt{3} u_0}{g} \]