A body is projected with a velocity u at an angle of `60^(@)` to the horizontal. The time interval after which it will be moving in a direction of `30^(@)` to the horizontal is

Options

A. `(u)/(sqrt3g)`
B. `(sqrt3u)/(g)`
C. `(sqrt3u)/(2g)`
D. `(2u)/(sqrt3u)`

Answer

To solve the problem, we need to find the time interval after which a body projected at an angle of \(60^\circ\) to the horizontal will be moving at an angle of \(30^\circ\) to the horizontal.

### Step-by-Step Solution:

1. **Identify the components of the initial velocity**:
The initial velocity \(u\) can be resolved into horizontal and vertical components.
- Horizontal component:
\[ u_x = u \cos(60^\circ) = u \cdot \frac{1}{2} = \frac{u}{2} \]
- Vertical component:
\[ u_y = u \sin(60^\circ) = u \cdot \frac{\sqrt{3}}{2} = \frac{u \sqrt{3}}{2} \]

2. **Write the equations of motion**:
The horizontal and vertical positions as functions of time \(t\) are given by:
- Horizontal position:
\[ x(t) = u_x \cdot t = \frac{u}{2} t \]
- Vertical position:
\[ y(t) = u_y \cdot t - \frac{1}{2} g t^2 = \frac{u \sqrt{3}}{2} t - \frac{1}{2} g t^2 \]

3. **Determine the condition for the angle of \(30^\circ\)**:
At time \(t\), the body will be at an angle of \(30^\circ\) to the horizontal when:
\[ \tan(30^\circ) = \frac{y(t)}{x(t)} \]
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), we can write:
\[ \frac{y(t)}{x(t)} = \frac{1}{\sqrt{3}} \]

4. **Substitute the expressions for \(y(t)\) and \(x(t)\)**:
\[ \frac{\frac{u \sqrt{3}}{2} t - \frac{1}{2} g t^2}{\frac{u}{2} t} = \frac{1}{\sqrt{3}} \]

5. **Simplify the equation**:
Cross-multiplying gives:
\[ \sqrt{3} \left(\frac{u \sqrt{3}}{2} t - \frac{1}{2} g t^2\right) = \frac{u}{2} t \]
This simplifies to:
\[ \frac{3u}{2} t - \frac{\sqrt{3}}{2} g t^2 = \frac{u}{2} t \]

6. **Rearranging the terms**:
\[ \frac{3u}{2} t - \frac{u}{2} t = \frac{\sqrt{3}}{2} g t^2 \]
\[ \frac{2u}{2} t = \frac{\sqrt{3}}{2} g t^2 \]
\[ ut = \frac{\sqrt{3}}{2} g t^2 \]

7. **Solving for \(t\)**:
Dividing both sides by \(t\) (assuming \(t \neq 0\)):
\[ u = \frac{\sqrt{3}}{2} g t \]
Rearranging gives:
\[ t = \frac{2u}{\sqrt{3} g} \]

### Final Answer:
The time interval after which the body will be moving in a direction of \(30^\circ\) to the horizontal is:
\[ t = \frac{2u}{\sqrt{3} g} \]