Two stones are projected from the top of a tower in opposite directions, with the same velocity V but at `30^(@)` & `60^(@)` with horizontal respectively. The relative velocity of first stone relative to second stone is

Options

A. 2 V
B. `sqrt2V`
C. `(2V)/(sqrt3)`
D. `(V)/(sqrt2)`

Answer

To solve the problem of finding the relative velocity of the first stone (projected at 30 degrees) relative to the second stone (projected at 60 degrees), we will follow these steps:

### Step 1: Resolve the velocities of both stones into components
Let’s denote:
- Stone A is projected at an angle of 30 degrees with the horizontal.
- Stone B is projected at an angle of 60 degrees with the horizontal.

For Stone A:
- Horizontal component of velocity (Vx_A) = V * cos(30°) = V * (√3/2)
- Vertical component of velocity (Vy_A) = V * sin(30°) = V * (1/2)

For Stone B:
- Horizontal component of velocity (Vx_B) = V * cos(60°) = V * (1/2)
- Vertical component of velocity (Vy_B) = V * sin(60°) = V * (√3/2)

### Step 2: Write the velocity vectors for both stones
The velocity vector for Stone A (VA) can be written as:
\[ \mathbf{V_A} = \left( V \cdot \frac{\sqrt{3}}{2}, V \cdot \frac{1}{2} \right) \]

The velocity vector for Stone B (VB) can be written as:
\[ \mathbf{V_B} = \left( -V \cdot \frac{1}{2}, V \cdot \frac{\sqrt{3}}{2} \right) \]
(Note: The horizontal component for Stone B is negative because it is projected in the opposite direction.)

### Step 3: Calculate the relative velocity of A with respect to B
The relative velocity of A with respect to B is given by:
\[ \mathbf{V_{AB}} = \mathbf{V_A} - \mathbf{V_B} \]

Substituting the values:
\[ \mathbf{V_{AB}} = \left( V \cdot \frac{\sqrt{3}}{2} - \left(-V \cdot \frac{1}{2}\right), V \cdot \frac{1}{2} - V \cdot \frac{\sqrt{3}}{2} \right) \]
\[ \mathbf{V_{AB}} = \left( V \cdot \frac{\sqrt{3}}{2} + V \cdot \frac{1}{2}, V \cdot \frac{1}{2} - V \cdot \frac{\sqrt{3}}{2} \right) \]

### Step 4: Simplify the components of the relative velocity
Now, simplifying the components:
1. Horizontal component:
\[ V_{x_{AB}} = V \left( \frac{\sqrt{3}}{2} + \frac{1}{2} \right) = V \left( \frac{\sqrt{3} + 1}{2} \right) \]

2. Vertical component:
\[ V_{y_{AB}} = V \left( \frac{1}{2} - \frac{\sqrt{3}}{2} \right) = V \left( \frac{1 - \sqrt{3}}{2} \right) \]

### Step 5: Calculate the magnitude of the relative velocity
The magnitude of the relative velocity vector can be calculated using the formula:
\[ |\mathbf{V_{AB}}| = \sqrt{(V_{x_{AB}})^2 + (V_{y_{AB}})^2} \]

Substituting the components:
\[ |\mathbf{V_{AB}}| = \sqrt{\left(V \cdot \frac{\sqrt{3} + 1}{2}\right)^2 + \left(V \cdot \frac{1 - \sqrt{3}}{2}\right)^2} \]

Factoring out \( V^2/4 \):
\[ |\mathbf{V_{AB}}| = \frac{V}{2} \sqrt{(\sqrt{3} + 1)^2 + (1 - \sqrt{3})^2} \]

### Step 6: Simplify the expression under the square root
Calculating:
1. \( (\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3} \)
2. \( (1 - \sqrt{3})^2 = 1 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3} \)

Adding these:
\[ 4 + 2\sqrt{3} + 4 - 2\sqrt{3} = 8 \]

Thus:
\[ |\mathbf{V_{AB}}| = \frac{V}{2} \sqrt{8} = \frac{V}{2} \cdot 2\sqrt{2} = V\sqrt{2} \]

### Final Result
The relative velocity of the first stone relative to the second stone is:
\[ V_{AB} = V\sqrt{2} \]