A man is walking due east at the rate of `2kmph`.The rain appears to him to come down vertically at the rate of `2kmph`.The actual velocity and direction of rainfall with the vertical respectively are

Options

A. `2sqrt2kmph,45^(@)`
B. `1/sqrt2kmph,30^(@)`
C. `2 kmph,0^(@)`
D. `1 kmph,90^(@)`

Explanation

To solve the problem, we need to determine the actual velocity and direction of the rainfall based on the given information.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A man is walking due east at a speed of 2 km/h.
- The rain appears to him to be falling vertically at a speed of 2 km/h.

2. **Setting Up the Vectors**:
- Let the velocity of the man (Vm) be represented as a vector pointing east (horizontally) with a magnitude of 2 km/h.
- Let the velocity of the rain with respect to the man (Vr_m) be represented as a vector pointing vertically downward with a magnitude of 2 km/h.

3. **Using Relative Velocity**:
- The actual velocity of the rain (Vr) can be found using the equation:
\[ Vr = Vr_m + Vm \]
- Here, \(Vr_m\) is the velocity of rain with respect to the man (2 km/h downward) and \(Vm\) is the velocity of the man (2 km/h eastward).

4. **Vector Addition**:
- We can visualize this as a right-angled triangle where:
- One side represents the velocity of the man (2 km/h east).
- The other side represents the velocity of the rain with respect to the man (2 km/h downward).
- To find the magnitude of the actual velocity of the rain, we use the Pythagorean theorem:
\[ |Vr| = \sqrt{(Vm)^2 + (Vr_m)^2} = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \text{ km/h} \]

5. **Finding the Direction**:
- The direction of the actual velocity of the rain can be found using the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{Vr_m}{Vm} = \frac{2}{2} = 1 \]
- Therefore, \(\theta = 45^\circ\) with respect to the vertical.

6. **Final Answer**:
- The actual velocity of the rain is \(2\sqrt{2} \text{ km/h}\) and it makes an angle of \(45^\circ\) with the vertical.

### Summary:
- Actual velocity of rainfall: \(2\sqrt{2} \text{ km/h}\)
- Direction of rainfall with respect to the vertical: \(45^\circ\)