Raindrops are falling with velocity `10 sqrt2` m/s making and angle `45^(@)` with the vertical. The drops appear to be falling vertically to a man running with constant velocity . The velocity of rain drops change such that the rain drops now appear to be falling vertically with `sqrt3` times the velcoity it appeared earlier to the same person running with same velocity .
After the velocity of rain drops change the magnitude of velocity of raindrops with respect to ground is

Options

A. `20sqrt3 m//s `
B. 10 m/s
C. `10 sqrt3 m//s`
D. 20 m/s

Answer

To solve the problem step by step, we need to analyze the motion of the raindrops and the man running.

### Step 1: Understand the initial conditions
The raindrops are falling with a velocity of \(10\sqrt{2}\) m/s at an angle of \(45^\circ\) with the vertical. We need to resolve this velocity into horizontal and vertical components.

**Hint:** Use trigonometric functions to resolve the velocity into components based on the given angle.

### Step 2: Resolve the velocity of raindrops
The vertical component \(V_{ry}\) and horizontal component \(V_{rx}\) of the raindrops can be calculated as follows:

- Vertical component:
\[ V_{ry} = -10\sqrt{2} \cdot \cos(45^\circ) = -10\sqrt{2} \cdot \frac{1}{\sqrt{2}} = -10 \text{ m/s} \]

- Horizontal component:
\[ V_{rx} = -10\sqrt{2} \cdot \sin(45^\circ) = -10\sqrt{2} \cdot \frac{1}{\sqrt{2}} = -10 \text{ m/s} \]

Thus, the velocity of the raindrops with respect to the ground is:
\[ \vec{V}_{RG} = -10\hat{i} - 10\hat{j} \text{ m/s} \]

**Hint:** Remember that the negative sign indicates the direction of the velocity components.

### Step 3: Determine the velocity of the man
Let the velocity of the man running be \(V\hat{i}\). Since the raindrops appear to fall vertically to the man, the horizontal component of the rain's velocity must equal the man's velocity:
\[ V + 10 = 0 \implies V = -10 \text{ m/s} \]

**Hint:** The horizontal component of the rain's velocity must cancel out with the man's velocity for the rain to appear vertical.

### Step 4: Analyze the new condition
Now, the problem states that the raindrops appear to fall vertically with a velocity that is \(\sqrt{3}\) times the previous vertical velocity. The previous vertical velocity was \(-10\hat{j}\). Therefore, the new vertical component of the rain's velocity \(V'_{ry}\) is:
\[ V'_{ry} = -10\sqrt{3} \text{ m/s} \]

The horizontal component remains unchanged:
\[ V'_{rx} = -10 \text{ m/s} \]

**Hint:** The vertical component changes, but the horizontal component remains the same.

### Step 5: Calculate the new velocity of the raindrops
The new velocity of the raindrops with respect to the ground is:
\[ \vec{V}'_{RG} = -10\hat{i} - 10\sqrt{3}\hat{j} \text{ m/s} \]

### Step 6: Find the magnitude of the new velocity
To find the magnitude of the velocity of the raindrops with respect to the ground, we use the Pythagorean theorem:
\[ |\vec{V}'_{RG}| = \sqrt{(-10)^2 + (-10\sqrt{3})^2} \]
\[ = \sqrt{100 + 300} = \sqrt{400} = 20 \text{ m/s} \]

**Hint:** Use the formula for the magnitude of a vector to combine the components.

### Final Answer
The magnitude of the velocity of the raindrops with respect to the ground after the change is \(20\) m/s.

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