A particle `A` is projected with speed `V_(A)` from a point making an angle `60^(@)` with the horizontal. At the same instant, a second particle `B` is thrown vertically upwards from a point directly below the maximum height point of parabolic path of A, with velocity `V_(B)` . If the two particles collide then the ratio of `V_(A)//V_(B)` should be:

Options

A. `1`
B. `2//sqrt(3)`
C. `sqrt(3)//2`
D. `sqrt(3)`

Explanation

To solve the problem, we need to analyze the motion of both particles A and B and find the ratio of their velocities when they collide.

### Step-by-Step Solution:

1. **Understanding the Motion of Particle A**:
- Particle A is projected with speed \( V_A \) at an angle of \( 60^\circ \) with the horizontal.
- The vertical component of the velocity of A can be calculated as:
\[ V_{Ay} = V_A \sin(60^\circ) = V_A \cdot \frac{\sqrt{3}}{2} \]

2. **Time Taken by Particle A to Reach Maximum Height**:
- The time taken by particle A to reach its maximum height can be calculated using the formula:
\[ t_A = \frac{V_{Ay}}{g} = \frac{V_A \cdot \frac{\sqrt{3}}{2}}{g} \]
- Simplifying this gives:
\[ t_A = \frac{V_A \sqrt{3}}{2g} \]

3. **Understanding the Motion of Particle B**:
- Particle B is thrown vertically upwards from a point directly below the maximum height of A with velocity \( V_B \).
- The time taken by particle B to reach its maximum height is given by:
\[ t_B = \frac{V_B}{g} \]

4. **Setting the Times Equal**:
- Since both particles collide at the same height at the same time, we set \( t_A \) equal to \( t_B \):
\[ \frac{V_A \sqrt{3}}{2g} = \frac{V_B}{g} \]
- We can cancel \( g \) from both sides:
\[ \frac{V_A \sqrt{3}}{2} = V_B \]

5. **Finding the Ratio \( \frac{V_A}{V_B} \)**:
- Rearranging the equation gives:
\[ V_A = \frac{2V_B}{\sqrt{3}} \]
- Therefore, the ratio \( \frac{V_A}{V_B} \) is:
\[ \frac{V_A}{V_B} = \frac{2}{\sqrt{3}} \]

### Final Answer:
The ratio of \( V_A \) to \( V_B \) is:
\[ \frac{V_A}{V_B} = \frac{2}{\sqrt{3}} \]