A stationary man observes that the rain is falling vertically downwards. When he starts running a velocity of `12 kmh^(-1)`, he observes that the rain is falling at an angle `60^(@)` with the vertical. The actual velocity of rain is

Options

A. `12 sqrt(3)kmh^(-1)`
B. `6 sqrt(3) kmh^(-1)`
C. `4 sqrt(3)kmh^(-1)`
D. `2sqrt(3)kmh^(-1)`

Explanation

To solve the problem, we need to analyze the situation using vector components and trigonometry.

### Step 1: Understand the scenario
A stationary observer sees the rain falling vertically. When he starts running at a speed of 12 km/h, he observes the rain falling at an angle of 60° with the vertical.

### Step 2: Define the velocities
- Let \( V_m \) be the velocity of the man = 12 km/h (to the right).
- Let \( V_r \) be the actual velocity of the rain (which we need to find).
- The observed velocity of the rain with respect to the man, \( V_{mr} \), makes an angle of 60° with the vertical.

### Step 3: Set up the relationship
From the problem, we know:
\[ V_{mr} = V_m - V_r \]
This means the velocity of the rain as observed by the man is the vector difference between the man's velocity and the actual velocity of the rain.

### Step 4: Break down the observed velocity into components
Since the rain is falling at an angle of 60° with the vertical, we can break it down into components:
- The vertical component of the observed velocity of rain is \( V_{mr} \cos(60°) \).
- The horizontal component of the observed velocity of rain is \( V_{mr} \sin(60°) \).

### Step 5: Use trigonometric relationships
From the angle:
- \( \cos(60°) = \frac{1}{2} \)
- \( \sin(60°) = \frac{\sqrt{3}}{2} \)

Thus, we can express the components as:
- Vertical component: \( V_{mr} \cdot \frac{1}{2} \)
- Horizontal component: \( V_{mr} \cdot \frac{\sqrt{3}}{2} \)

### Step 6: Relate the components to the man's velocity
Since the rain appears to fall at an angle of 60° with the vertical, we can equate the horizontal component of the observed velocity to the man's velocity:
\[ V_m = V_{mr} \cdot \frac{\sqrt{3}}{2} \]

### Step 7: Substitute the known values
Substituting \( V_m = 12 \) km/h:
\[ 12 = V_{mr} \cdot \frac{\sqrt{3}}{2} \]

### Step 8: Solve for \( V_{mr} \)
Rearranging gives:
\[ V_{mr} = \frac{12 \cdot 2}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3} \text{ km/h} \]

### Step 9: Find the actual velocity of the rain
Now, we can find the actual velocity of the rain using the vertical component:
\[ V_r = V_{mr} \cdot \cos(60°) \]
\[ V_r = 8\sqrt{3} \cdot \frac{1}{2} = 4\sqrt{3} \text{ km/h} \]

### Final Answer
The actual velocity of the rain is \( 4\sqrt{3} \) km/h.

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