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 m s^-1`
B. `1.0 m s^-1`
C. `0.5 (sqrt(3)) m s^-1`
D. `0.43 m s^-1`

Explanation

To solve the problem, we need to find the speed of the rain with respect to the stationary man. Let's break down the solution step by step.

### Step 1: Understand the scenario
A stationary man observes rain falling at an angle of \(30^\circ\) with the vertical. This means that the rain has a horizontal component of velocity relative to the man.

### Step 2: Define the velocities
Let:
- \(V_r\) = speed of the rain (magnitude)
- \(V_m = 0.5 \, \text{m/s}\) = speed of the man moving forward

### Step 3: Resolve the rain's velocity into components
The rain's velocity can be resolved into two components:
- Vertical component: \(V_r \cos(30^\circ)\)
- Horizontal component: \(V_r \sin(30^\circ)\)

### Step 4: Set up the equation for the stationary man
For the stationary man, the horizontal component of the rain's velocity must equal the man's velocity when he starts moving. Thus, we have:
\[ V_r \sin(30^\circ) = V_m \]
Substituting \(V_m = 0.5 \, \text{m/s}\) and \(\sin(30^\circ) = \frac{1}{2}\):
\[ V_r \cdot \frac{1}{2} = 0.5 \]
\[ V_r = 1 \, \text{m/s} \]

### Step 5: Find the velocity of the rain with respect to the man
Now, we need to find the speed of the rain with respect to the moving man. The relative velocity \(V_{rm}\) can be calculated using the Pythagorean theorem:
\[ V_{rm} = \sqrt{V_r^2 + V_m^2} \]
Substituting the values we found:
\[ V_{rm} = \sqrt{(1)^2 + (0.5)^2} \]
\[ V_{rm} = \sqrt{1 + 0.25} = \sqrt{1.25} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \, \text{m/s} \]

### Step 6: Final answer
Thus, the speed of rain with respect to the stationary man is:
\[ V_{rm} = \frac{\sqrt{5}}{2} \, \text{m/s} \]