To a stationary man, rain appears to be falling at his back at an angle `30^@` with the vertical. As he starts moving forward with a speed of `0.5 m s^-1`, he finds that the rain is falling vertically.
The speed of rain with respect to the stationary man is.

Options

A. `0.5 ms^(-1)`
B. `1 ms ^(-1)`
C. `sqrt(3)/2 ms ^(-1)`
D. `1/sqrt3 ms ^(-1)`

Answer

To solve the problem, we need to analyze the situation using the concept of relative velocity. Let's break it down step by step.

### Step 1: Understand the Situation
A stationary man observes the rain falling at an angle of \(30^\circ\) with the vertical. This means that the rain has both vertical and horizontal components of velocity.

### Step 2: Define the Velocities
Let:
- \(v_r\) = speed of rain
- The angle of rain with the vertical = \(30^\circ\)

From the angle, we can derive the components of the rain's velocity:
- Vertical component of rain's velocity, \(v_{r_y} = v_r \cos(30^\circ)\)
- Horizontal component of rain's velocity, \(v_{r_x} = v_r \sin(30^\circ)\)

### Step 3: Analyze the Man's Movement
When the man starts moving forward with a speed of \(0.5 \, \text{m/s}\), he observes the rain falling vertically. This means that the horizontal component of the rain's velocity must equal the man's speed.

### Step 4: Set Up the Equation
For the rain to appear to fall vertically to the man, the horizontal component of the rain's velocity must equal the man's speed:
\[ v_{r_x} = v_{man} \]
Substituting the values:
\[ v_r \sin(30^\circ) = 0.5 \, \text{m/s} \]

### Step 5: Solve for the Speed of Rain
Now, we can solve for \(v_r\):
\[ v_r \cdot \frac{1}{2} = 0.5 \]
\[ v_r = 0.5 \cdot 2 = 1 \, \text{m/s} \]

### Conclusion
The speed of the rain with respect to the stationary man is \(1 \, \text{m/s}\).

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