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

2. **Finding the Maximum Height of Particle A**:
- The maximum height \( H \) attained by particle A can be calculated using the formula:
\[ H = \frac{V_{Ay}^2}{2g} = \frac{\left(V_A \cdot \frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{3V_A^2}{8g} \]

3. **Understanding the Motion of Particle B**:
- Particle B is thrown vertically upwards from a point directly below the maximum height of particle A with an initial velocity \( V_B \).
- The height reached by particle B when it collides with particle A is also \( H \).

4. **Using the Equations of Motion for Particle B**:
- The height \( H \) reached by particle B can be expressed as:
\[ H = V_B t - \frac{1}{2} g t^2 \]
- Setting the height \( H \) from both particles equal gives:
\[ V_B t - \frac{1}{2} g t^2 = \frac{3V_A^2}{8g} \]

5. **Using the Time of Flight for Particle A**:
- The time \( t \) taken by particle A to reach the maximum height can be calculated using:
\[ t = \frac{V_{Ay}}{g} = \frac{V_A \cdot \frac{\sqrt{3}}{2}}{g} = \frac{V_A \sqrt{3}}{2g} \]

6. **Substituting Time into the Equation for Particle B**:
- Substitute \( t \) into the height equation for particle B:
\[ H = V_B \left(\frac{V_A \sqrt{3}}{2g}\right) - \frac{1}{2} g \left(\frac{V_A \sqrt{3}}{2g}\right)^2 \]
- Simplifying gives:
\[ H = \frac{V_B V_A \sqrt{3}}{2g} - \frac{3V_A^2}{8g} \]

7. **Equating the Heights**:
- Set the two expressions for \( H \) equal to each other:
\[ \frac{3V_A^2}{8g} = \frac{V_B V_A \sqrt{3}}{2g} - \frac{3V_A^2}{8g} \]
- Rearranging leads to:
\[ \frac{3V_A^2}{4g} = \frac{V_B V_A \sqrt{3}}{2g} \]

8. **Finding the Ratio of Velocities**:
- Cancel \( g \) and \( V_A \) (assuming \( V_A \neq 0 \)):
\[ \frac{3V_A}{4} = \frac{V_B \sqrt{3}}{2} \]
- Rearranging gives:
\[ \frac{V_A}{V_B} = \frac{2}{\sqrt{3}} \]

### Final Ratio:
Thus, the ratio of \( V_A \) to \( V_B \) is:
\[ \frac{V_A}{V_B} = \frac{2}{\sqrt{3}} \]