A particle `A` is projected with speed `V_(A)` from a point making an angle `60^(@)` with the horizontal. At the same instant, second particle `B(` lie in the same horizontal plane `)` 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. **Identify the Components of Velocity for Particle A:**
- Particle A is projected at an angle of \(60^\circ\) with a speed \(V_A\).
- The vertical component of the velocity \(V_{Ay}\) is given by:
\[ V_{Ay} = V_A \sin(60^\circ) = V_A \cdot \frac{\sqrt{3}}{2} \]
- The horizontal component of the velocity \(V_{Ax}\) is:
\[ V_{Ax} = V_A \cos(60^\circ) = V_A \cdot \frac{1}{2} \]

2. **Vertical Displacement of Particle A:**
- The vertical displacement \(S_{yA}\) of particle A after time \(t\) can be expressed using the equation of motion:
\[ S_{yA} = V_{Ay} \cdot t - \frac{1}{2} g t^2 \]
- Substituting \(V_{Ay}\):
\[ S_{yA} = \left(V_A \cdot \frac{\sqrt{3}}{2}\right) t - \frac{1}{2} g t^2 \]

3. **Vertical Displacement of Particle B:**
- Particle B is thrown vertically upwards from the point directly below the maximum height of A.
- The vertical displacement \(S_{yB}\) of particle B after time \(t\) is:
\[ S_{yB} = V_B \cdot t - \frac{1}{2} g t^2 \]

4. **Setting the Displacements Equal:**
- Since both particles collide, their vertical displacements must be equal:
\[ S_{yA} = S_{yB} \]
- Therefore, we have:
\[ \left(V_A \cdot \frac{\sqrt{3}}{2}\right) t - \frac{1}{2} g t^2 = V_B \cdot t - \frac{1}{2} g t^2 \]

5. **Canceling Common Terms:**
- The \(-\frac{1}{2} g t^2\) terms cancel out:
\[ \left(V_A \cdot \frac{\sqrt{3}}{2}\right) t = V_B \cdot t \]

6. **Dividing by \(t\) (assuming \(t \neq 0\)):**
- We can simplify this to:
\[ V_A \cdot \frac{\sqrt{3}}{2} = V_B \]

7. **Finding the Ratio \( \frac{V_A}{V_B} \):**
- Rearranging gives us:
\[ \frac{V_A}{V_B} = \frac{2}{\sqrt{3}} \]

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