A man walking with a speed of 3 km/h finds the rain drops falling vertically downwards. When the man increases his speed to 6km/h he find that the rain drops are falling making an angle of `30^(@)` with the vertical . Find the speed of the rain drops ( in km/h)

Answer

To solve the problem, we will analyze the motion of the man and the rain using vector components.

### Step-by-Step Solution:

1. **Understanding the Situation**:
- When the man walks at 3 km/h, he sees the rain falling vertically. This means that the horizontal component of the rain's velocity must equal the man's velocity.
- When the man increases his speed to 6 km/h, he observes the rain making a 30-degree angle with the vertical.

2. **Setting Up the Vectors**:
- Let \( V_r \) be the speed of the rain (which we need to find).
- The velocity of the man when walking at 3 km/h is \( \vec{V_m} = 3 \hat{i} \) (to the right).
- The velocity of the rain with respect to the ground can be broken down into components:
- Horizontal component: \( V_r \sin(\theta) \)
- Vertical component: \( V_r \cos(\theta) \)

3. **First Case (Man at 3 km/h)**:
- Since the man sees the rain falling vertically, the horizontal component of the rain's velocity must equal the man's speed:
\[ V_r \sin(90^\circ) = 3 \quad \Rightarrow \quad V_r \sin(90^\circ) = 3 \quad \Rightarrow \quad V_r = 3 \quad \text{(This is not used directly, just a reference)} \]

4. **Second Case (Man at 6 km/h)**:
- Now the man is walking at 6 km/h and sees the rain making a 30-degree angle with the vertical. The horizontal component of the rain's velocity is:
\[ V_r \sin(30^\circ) = 6 \]
- Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ V_r \cdot \frac{1}{2} = 6 \quad \Rightarrow \quad V_r = 12 \text{ km/h} \]

5. **Finding the Vertical Component**:
- The vertical component of the rain's velocity is:
\[ V_r \cos(30^\circ) = V_r \cdot \frac{\sqrt{3}}{2} \]
- Therefore, the vertical component is:
\[ V_r \cdot \frac{\sqrt{3}}{2} = 12 \cdot \frac{\sqrt{3}}{2} = 6\sqrt{3} \text{ km/h} \]

6. **Final Result**:
- The speed of the rain can be calculated using the Pythagorean theorem:
\[ V = \sqrt{(V_r \sin(30^\circ))^2 + (V_r \cos(30^\circ))^2} \]
- Substituting the values:
\[ V = \sqrt{(6)^2 + (6\sqrt{3})^2} = \sqrt{36 + 108} = \sqrt{144} = 12 \text{ km/h} \]

### Final Answer:
The speed of the rain drops is **12 km/h**.

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