Rain drops are falling vertically down wards at `5sqrt(2)` m/s. A man runs horizontally in the rain at `5sqrt(2)`. The magnitude and direction of relative velocity of the rain drops with respect to the person is

Options

A. `5 m//s , 45^(@)` with vertical
B. `10 m//s, 45^(@)` with vertical
C. `15 m//s, 30^(@)` with vertical
D. None of these

Answer

To solve the problem of finding the magnitude and direction of the relative velocity of the raindrops with respect to the person, we will follow these steps:

### Step 1: Identify the velocities
- The velocity of the raindrops (Vr) is given as \( 5\sqrt{2} \) m/s vertically downward.
- The velocity of the man (Vp) is given as \( 5\sqrt{2} \) m/s horizontally.

### Step 2: Set up the relative velocity equation
The relative velocity of the rain with respect to the person (Vr/P) can be expressed as:
\[ \vec{V_{r/p}} = \vec{V_r} - \vec{V_p} \]
Since the man is running horizontally, we can represent the velocities in vector form:
- \( \vec{V_r} = (0, -5\sqrt{2}) \) m/s (downward)
- \( \vec{V_p} = (5\sqrt{2}, 0) \) m/s (horizontal)

### Step 3: Calculate the relative velocity vector
Now, we can calculate the relative velocity:
\[ \vec{V_{r/p}} = (0, -5\sqrt{2}) - (5\sqrt{2}, 0) = (-5\sqrt{2}, -5\sqrt{2}) \]

### Step 4: Find the magnitude of the relative velocity
To find the magnitude of the relative velocity, we use the Pythagorean theorem:
\[ |\vec{V_{r/p}}| = \sqrt{(-5\sqrt{2})^2 + (-5\sqrt{2})^2} \]
Calculating this gives:
\[ |\vec{V_{r/p}}| = \sqrt{(25 \cdot 2) + (25 \cdot 2)} = \sqrt{50 + 50} = \sqrt{100} = 10 \text{ m/s} \]

### Step 5: Find the direction of the relative velocity
To find the angle (θ) that the relative velocity makes with the vertical, we can use the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{V_{p}}{V_{r}} = \frac{5\sqrt{2}}{5\sqrt{2}} = 1 \]
Thus,
\[ \theta = \tan^{-1}(1) = 45^\circ \]

### Conclusion
The magnitude of the relative velocity of the raindrops with respect to the person is \( 10 \) m/s, and the direction is \( 45^\circ \) with respect to the vertical.

### Final Answer
The correct option is option 2: \( 10 \) m/s at \( 45^\circ \) with the vertical.

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