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 solve the problem of finding the time at which the velocities of two particles projected at angles of \(30^\circ\) and \(60^\circ\) with the horizontal become parallel, we can follow these steps:

### Step 1: Determine the velocity components of each particle
For both particles, the initial speed is \(V_0\).

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

**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 functions of time
The velocities of the particles at time \(t\) can be expressed as:

**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 conditions for parallel velocities
The velocities will be parallel when the ratio of their components is equal:
\[ \frac{V_{1x}}{V_{2x}} = \frac{V_{1y}}{V_{2y}} \]

Substituting the components:
\[ \frac{V_0 \cdot \frac{\sqrt{3}}{2}}{V_0 \cdot \frac{1}{2}} = \frac{V_0 \cdot \frac{1}{2} - gt}{V_0 \cdot \frac{\sqrt{3}}{2} - gt} \]

### Step 4: Simplify the equation
Cancel \(V_0\) from both sides:
\[ \frac{\sqrt{3}}{1} = \frac{\frac{1}{2} - gt}{\frac{\sqrt{3}}{2} - gt} \]

Cross-multiplying gives:
\[ \sqrt{3} \left(\frac{\sqrt{3}}{2} - gt\right) = \frac{1}{2} - gt \]

### Step 5: Expand and collect terms
Expanding the left side:
\[ \frac{3}{2} - \sqrt{3}gt = \frac{1}{2} - gt \]

Rearranging gives:
\[ \frac{3}{2} - \frac{1}{2} = gt - \sqrt{3}gt \]
\[ 1 = gt(1 - \sqrt{3}) \]

### Step 6: Solve for time \(t\)
\[ t = \frac{1}{g(1 - \sqrt{3})} \]

### Step 7: Rationalize the denominator
To rationalize:
\[ t = \frac{1(1 + \sqrt{3})}{g(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{1 + \sqrt{3}}{g(1 - 3)} = \frac{1 + \sqrt{3}}{-2g} \]
Thus,
\[ t = \frac{(1 + \sqrt{3})}{2g} \]

### Final Answer
The time at which the velocities of the two particles become parallel is:
\[ t = \frac{(1 + \sqrt{3}) V_0}{2g} \]