Rain is falling vertically downwards with a speed of `4 km h^-1`. A girl moves on a straight road with a velocity of `3 km h^-1`. The apparent velocity of rain with respect to the girl is.

Options

A. `3 km h^-1`
B. `4 km h^-1`
C. 5 km h^-1`
D. `7 km h^-1`

Explanation

To find the apparent velocity of rain with respect to the girl, we can follow these steps:

### Step 1: Understand the velocities involved
- The rain is falling vertically downwards with a speed of \( V_r = 4 \, \text{km/h} \).
- The girl is moving horizontally on a straight road with a speed of \( V_g = 3 \, \text{km/h} \).

### Step 2: Set up the vector representation
- We can represent the velocity of the rain as a vector pointing downward:
\[ \vec{V_r} = 0 \hat{i} - 4 \hat{j} \, \text{km/h} \]
- The velocity of the girl can be represented as a vector pointing horizontally:
\[ \vec{V_g} = 3 \hat{i} + 0 \hat{j} \, \text{km/h} \]

### Step 3: Calculate the apparent velocity of rain with respect to the girl
The apparent velocity of the rain with respect to the girl is given by:
\[ \vec{V_{rg}} = \vec{V_r} - \vec{V_g} \]
Substituting the vectors:
\[ \vec{V_{rg}} = (0 \hat{i} - 4 \hat{j}) - (3 \hat{i} + 0 \hat{j}) = -3 \hat{i} - 4 \hat{j} \, \text{km/h} \]

### Step 4: Find the magnitude of the resultant vector
To find the magnitude of the apparent velocity, we use the Pythagorean theorem:
\[ |\vec{V_{rg}}| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{km/h} \]

### Step 5: Find the direction of the apparent velocity
To find the angle \( \theta \) with respect to the vertical, we use the tangent function:
\[ \tan \theta = \frac{|\text{horizontal component}|}{|\text{vertical component}|} = \frac{3}{4} \]
Thus,
\[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \]

### Final Result
The apparent velocity of the rain with respect to the girl is \( 5 \, \text{km/h} \) at an angle \( \theta = \tan^{-1}\left(\frac{3}{4}\right) \) with respect to the vertical.

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