A car is moving on horizontal ground with a velocity vm `s^(-1)` with respect to earth. The driver of the car observes the rain to be falling vertically downwards with velocity 2vm `s^(-1)` . What is the velocity of the rain with respect to the ground ?

Options

A. 3v
B. `vsqrt(3)`
C. `vsqrt(5)`
D. none of these

Explanation

To solve the problem, we need to determine the velocity of the rain with respect to the ground given the velocity of the car and the observed velocity of the rain by the driver.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Velocity of the car with respect to the ground, \( V_c = v \) (in the horizontal direction).
- Velocity of the rain with respect to the car, \( V_{r/c} = 2v \) (in the vertical downward direction).

2. **Understand the Vector Relationship:**
- The velocity of the rain with respect to the ground can be expressed using vector addition:
\[ V_{r/g} = V_{r/c} + V_c \]
- Here, \( V_{r/g} \) is the velocity of the rain with respect to the ground, \( V_{r/c} \) is the velocity of the rain with respect to the car, and \( V_c \) is the velocity of the car with respect to the ground.

3. **Set Up the Vectors:**
- Let the horizontal direction (the direction of the car's motion) be the x-axis, and the vertical direction (the direction of the rain) be the y-axis.
- Therefore, we can represent the vectors as:
- \( V_c = v \hat{i} \) (horizontal)
- \( V_{r/c} = -2v \hat{j} \) (downward)

4. **Calculate the Resultant Velocity:**
- The resultant velocity of the rain with respect to the ground can be calculated using the Pythagorean theorem since the two vectors are perpendicular:
\[ V_{r/g} = \sqrt{(V_c)^2 + (V_{r/c})^2} \]
- Substituting the values:
\[ V_{r/g} = \sqrt{(v)^2 + (-2v)^2} \]
\[ = \sqrt{v^2 + 4v^2} \]
\[ = \sqrt{5v^2} \]
\[ = v\sqrt{5} \]

5. **Conclusion:**
- The velocity of the rain with respect to the ground is:
\[ V_{r/g} = v\sqrt{5} \text{ m/s} \]

### Final Answer:
The velocity of the rain with respect to the ground is \( v\sqrt{5} \) m/s.

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