A man running at a speed of 5 km/h finds that the rain is falling vertically. When the stops running, the finds that the rain is falling at an angle of `60^(@)` with the horizontal. The velocity of rain with respect to running man is

Options

A. `(5)/(sqrt3)` km/h
B. `(5sqrt3)/(2)` km/h
C. `(4sqrt3)/(5)` km/h
D. `5sqrt3` km/h

Answer

To solve the problem step by step, we will analyze the situation using relative velocity concepts.

### Step 1: Understand the scenario
The man is running at a speed of 5 km/h. When he is running, he perceives the rain to be falling vertically. When he stops, he sees the rain falling at an angle of 60 degrees with the horizontal.

### Step 2: Define the velocities
Let:
- \( V_m = 5 \) km/h (velocity of the man)
- \( V_r \) = velocity of the rain (unknown)
- When the man is running, the rain appears to fall vertically, which means the horizontal component of the rain's velocity must equal the man's speed.

### Step 3: Analyze the situation when the man is running
When the man is running, the rain appears to fall vertically. This means that the horizontal component of the rain's velocity is equal to the man's speed:
\[ V_r \cos(60^\circ) = V_m \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ V_r \cdot \frac{1}{2} = 5 \]
\[ V_r = 10 \text{ km/h} \]

### Step 4: Analyze the situation when the man stops
When the man stops, the rain is falling at an angle of 60 degrees with the horizontal. We can use the tangent of the angle to find the relationship between the vertical and horizontal components of the rain's velocity:
\[ \tan(60^\circ) = \frac{V_r \sin(60^\circ)}{V_r \cos(60^\circ)} \]
Since \( \tan(60^\circ) = \sqrt{3} \):
\[ \sqrt{3} = \frac{V_r \cdot \frac{\sqrt{3}}{2}}{V_r \cdot \frac{1}{2}} \]
This simplifies to:
\[ \sqrt{3} = \frac{\sqrt{3}}{1} \]
This confirms that the angle is consistent with the velocities.

### Step 5: Calculate the relative velocity of rain with respect to the man
Now we can find the relative velocity of the rain with respect to the man using the formula:
\[ V_{rm} = \sqrt{V_r^2 + V_m^2 - 2V_r V_m \cos(60^\circ)} \]
Substituting the known values:
\[ V_{rm} = \sqrt{(10)^2 + (5)^2 - 2 \cdot 10 \cdot 5 \cdot \frac{1}{2}} \]
\[ = \sqrt{100 + 25 - 50} \]
\[ = \sqrt{75} \]
\[ = 5\sqrt{3} \text{ km/h} \]

### Final Answer
The velocity of the rain with respect to the running man is \( 5\sqrt{3} \) km/h.

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