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. Let's break it down step by step.

### Step 1: Understand the Situation
A stationary man observes the rain falling vertically downwards. This means that the velocity of the rain (Vr) when the man is stationary is purely vertical.

### Step 2: Introduce the Man's Motion
When the man starts running with a velocity of 12 km/h, he observes the rain falling at an angle of 60 degrees with the vertical. This indicates that the rain has both vertical and horizontal components when observed from the man's frame of reference.

### Step 3: Set Up the Components
1. Let the velocity of the rain be \( V_r \) (which has both vertical and horizontal components).
2. The horizontal component of the rain's velocity (when observed by the man) can be represented as \( V_{r,x} \).
3. The vertical component remains \( V_{r,y} \).

Since the man is running horizontally, the horizontal component of the rain's velocity relative to him is:
\[ V_{r,x} = V_m = 12 \text{ km/h} \]

### Step 4: Use the Angle Information
The angle of the rain with the vertical is given as 60 degrees. This means we can use trigonometric relationships to express the components of the rain's velocity.

Using the tangent of the angle:
\[ \tan(60^\circ) = \frac{V_{r,x}}{V_{r,y}} \]
We know that \( \tan(60^\circ) = \sqrt{3} \). Therefore:
\[ \sqrt{3} = \frac{12}{V_{r,y}} \]

### Step 5: Solve for the Vertical Component
Rearranging the equation gives us:
\[ V_{r,y} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \text{ km/h} \]

### Step 6: Calculate the Actual Velocity of Rain
Now, we can find the actual velocity of the rain using the Pythagorean theorem, since we have both components:
\[ V_r = \sqrt{V_{r,x}^2 + V_{r,y}^2} \]
Substituting the values we found:
\[ V_r = \sqrt{(12)^2 + (4\sqrt{3})^2} \]
Calculating each term:
\[ = \sqrt{144 + 48} = \sqrt{192} = 8\sqrt{3} \text{ km/h} \]

### Final Answer
The actual velocity of the rain is \( 8\sqrt{3} \text{ km/h} \).

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