Two particles are projected simultaneously with same speed `V_0` in same vertical plane at angle `30^@ and 60^@` With the horizontal .The time at which their velocities becomes parallel is

Options

A. `V_0/(2g)`
B. `V_0((sqrt(3)+1)/(2g))`
C. `(sqrt(3)V_0)/(2g)`
D. `(V_0(sqrt(3)-1))/g`

Answer

To find the time at which the velocities of two particles projected at angles \(30^\circ\) and \(60^\circ\) become parallel, we can follow these steps:

### Step 1: Write the velocity components for both particles
The initial velocity \(V_0\) can be broken down into horizontal and vertical components for both particles.

For Particle 1 (projected at \(30^\circ\)):
- Horizontal component:
\[ V_{1x} = V_0 \cos(30^\circ) = V_0 \cdot \frac{\sqrt{3}}{2} \]
- Vertical component:
\[ V_{1y} = V_0 \sin(30^\circ) = V_0 \cdot \frac{1}{2} \]

For Particle 2 (projected at \(60^\circ\)):
- Horizontal component:
\[ V_{2x} = V_0 \cos(60^\circ) = V_0 \cdot \frac{1}{2} \]
- Vertical component:
\[ V_{2y} = V_0 \sin(60^\circ) = V_0 \cdot \frac{\sqrt{3}}{2} \]

### Step 2: Write the velocity equations as a function of time
The velocities of the particles at time \(t\) can be expressed as follows, considering the effect of gravitational acceleration \(g\) acting downwards:

For Particle 1:
\[ \vec{V_1}(t) = \left(V_0 \cdot \frac{\sqrt{3}}{2}\right) \hat{i} + \left(V_0 \cdot \frac{1}{2} - gt\right) \hat{j} \]

For Particle 2:
\[ \vec{V_2}(t) = \left(V_0 \cdot \frac{1}{2}\right) \hat{i} + \left(V_0 \cdot \frac{\sqrt{3}}{2} - gt\right) \hat{j} \]

### Step 3: Set the velocities to be parallel
The velocities are parallel when the direction of the velocity vectors is the same. This means that the ratio of the vertical and horizontal components of both velocities must be equal:

\[ \frac{V_{1y}}{V_{1x}} = k \cdot \frac{V_{2y}}{V_{2x}} \]

Where \(k\) is some constant.

### Step 4: Substitute the components
Substituting the components we found earlier:

\[ \frac{V_0 \cdot \frac{1}{2} - gt}{V_0 \cdot \frac{\sqrt{3}}{2}} = k \cdot \frac{V_0 \cdot \frac{\sqrt{3}}{2} - gt}{V_0 \cdot \frac{1}{2}} \]

### Step 5: Simplify the equation
Since \(V_0\) is common in all terms, we can cancel it out:

\[ \frac{\frac{1}{2} - \frac{gt}{V_0}}{\frac{\sqrt{3}}{2}} = k \cdot \frac{\frac{\sqrt{3}}{2} - \frac{gt}{V_0}}{\frac{1}{2}} \]

### Step 6: Solve for \(t\)
Cross-multiplying gives us:

\[ \left(\frac{1}{2} - \frac{gt}{V_0}\right) = k \cdot \left(\frac{\sqrt{3}}{2} - \frac{gt}{V_0}\right) \cdot \frac{\sqrt{3}}{1} \]

From the horizontal components, we can find \(k\):

\[ \frac{\sqrt{3}}{2} = k \cdot \frac{1}{2} \]
\[ k = \sqrt{3} \]

Now substituting \(k\) back into the equation and solving for \(t\):

\[ \frac{1}{2} - \frac{gt}{V_0} = \sqrt{3} \left(\frac{\sqrt{3}}{2} - \frac{gt}{V_0}\right) \]

This leads to:

\[ \frac{1}{2} - \frac{gt}{V_0} = \frac{3}{2} - \sqrt{3} \cdot \frac{gt}{V_0} \]

Rearranging gives:

\[ \frac{gt}{V_0}(\sqrt{3} - 1) = 1 \]

Thus, we find:

\[ t = \frac{V_0}{g(\sqrt{3} - 1)} \]

### Final Answer
The time at which their velocities become parallel is:
\[ t = \frac{V_0}{g(\sqrt{3} - 1)} \]