A man running with velocity `3m//s` finds that raindrops are hitting him vertically with a speed of `4m//s`.
Magnitude of velocity of the raindrops is

Options

A. `3 m//s`
B. `4 m//s`
C. `5 m//s`
D. `6 m//s`

Explanation

To find the magnitude of the velocity of the raindrops, we can use vector addition. The raindrops are observed to be falling vertically downwards with a speed of 4 m/s relative to the man, who is running horizontally at a speed of 3 m/s.

### Step-by-step Solution:

1. **Identify the velocities**:
- The velocity of the man (VM) is given as 3 m/s in the horizontal direction (let's assume this is along the positive x-axis).
- The velocity of the raindrops with respect to the man (VR|M) is given as 4 m/s vertically downward (which we can represent as -4 m/s in the vertical direction, or negative y-axis).

2. **Set up the vector equation**:
- We can express the velocity of the raindrops (VR) in terms of the velocity of the man and the relative velocity of the rain.
- The equation is:
\[ \text{VR} = \text{VR|M} + \text{VM} \]
- In vector form, this can be written as:
\[ \text{VR} = -4 \hat{j} + 3 \hat{i} \]
- Here, \( \hat{i} \) represents the horizontal direction and \( \hat{j} \) represents the vertical direction.

3. **Calculate the magnitude of the velocity of the raindrops**:
- The magnitude of a vector \( \mathbf{V} = a \hat{i} + b \hat{j} \) is given by:
\[ |\mathbf{V}| = \sqrt{a^2 + b^2} \]
- In our case:
- \( a = 3 \) (horizontal component)
- \( b = -4 \) (vertical component)
- Therefore, the magnitude of the velocity of the raindrops is:
\[ |\text{VR}| = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ m/s} \]

4. **Conclusion**:
- The magnitude of the velocity of the raindrops is **5 m/s**.