To a man walking at the rate of `3 km//h` the rain appear to fall vertically downwards. When he increases his speed `6 km//h` it appears to meet him at an angle of `45^@` with vertical. Find the speed of rain.

Options

A. `sqrt2 km//h`
B. `2sqrt2 km//h`
C. `3sqrt2 km//h`
D. `5sqrt2 km//h`

Explanation

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the initial conditions
The man walks at a speed of \(3 \text{ km/h}\) and perceives the rain to fall vertically downwards. This means that the horizontal component of the rain's velocity must be equal to the man's speed, which is \(3 \text{ km/h}\).

### Step 2: Set up the equations
Let the speed of the rain be \(V_r\) (which we need to find). When the man walks at \(3 \text{ km/h}\), the velocity of the rain relative to him can be expressed as:
\[ \text{Velocity of rain relative to man} = \text{Velocity of rain} - \text{Velocity of man} \]
In vector form, this is:
\[ \vec{V}_{rm} = \vec{V}_r - 3\hat{i} = -V_r\hat{j} \]
Since the rain appears to fall vertically, the horizontal component must be zero, leading to:
\[ 3\hat{i} - V_r\hat{j} = 0 \]

### Step 3: Analyze the second condition
When the man increases his speed to \(6 \text{ km/h}\), the rain appears to meet him at an angle of \(45^\circ\) with the vertical. This means the horizontal and vertical components of the rain's velocity relative to the man are equal.

The velocity of the man now is \(6 \text{ km/h}\), so:
\[ \vec{V}_{rm} = \vec{V}_r - 6\hat{i} \]
At \(45^\circ\), the components satisfy:
\[ \text{Vertical component} = \text{Horizontal component} \]
Thus:
\[ -V_r = 6 \quad \text{(horizontal component)} \]
This implies:
\[ V_r = 6 \]

### Step 4: Solve for the speed of rain
Now, we have two equations:
1. From the first condition: \(V_r = 3\)
2. From the second condition: \(V_r = 6\)

However, we need to find the resultant speed of the rain. The resultant velocity of the rain can be expressed as:
\[ \vec{V}_r = 3\hat{i} - V_r\hat{j} \]
To find the magnitude of the rain's velocity:
\[ |\vec{V}_r| = \sqrt{(3)^2 + (V_r)^2} \]
Substituting \(V_r = 3\):
\[ |\vec{V}_r| = \sqrt{(3)^2 + (3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \text{ km/h} \]

### Final Answer
The speed of the rain is \(3\sqrt{2} \text{ km/h}\).

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