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 situation step by step.

### Step 1: Understand the situation
The man is walking with a speed of 3 km/h and perceives the rain falling vertically. This means that the horizontal component of the rain's velocity must be equal to the man's speed.

### Step 2: Establish the velocity of the man
Let:
- \( V_m = 3 \) km/h (initial speed of the man)
- The velocity of the rain with respect to the man is purely vertical, which means the horizontal component of the rain's velocity is equal to the man's speed.

### Step 3: Analyze the second scenario
When the man increases his speed to 6 km/h, he observes the rain making an angle of 30 degrees with the vertical.

Let:
- \( V_r \) = speed of the rain (unknown)
- The horizontal component of the rain's velocity remains the same.

### Step 4: Set up the equations
From the first scenario:
- Horizontal component of rain's velocity = \( V_m = 3 \) km/h

From the second scenario:
- The man’s new speed \( V_m' = 6 \) km/h
- The angle \( \theta = 30^\circ \)

Using trigonometry, we can express the vertical and horizontal components of the rain's velocity when the man is moving at 6 km/h:
- Horizontal component of rain's velocity = \( V_r \cos(30^\circ) \)
- Vertical component of rain's velocity = \( V_r \sin(30^\circ) \)

### Step 5: Set up the equations based on the angle
At \( 30^\circ \):
- The horizontal component of the rain's velocity is \( V_r \cos(30^\circ) = 3 \) km/h
- The vertical component of the rain's velocity is \( V_r \sin(30^\circ) \)

### Step 6: Solve for \( V_r \)
Using the cosine and sine values:
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)
- \( \sin(30^\circ) = \frac{1}{2} \)

From the horizontal component:
\[ V_r \cdot \frac{\sqrt{3}}{2} = 3 \]
\[ V_r = \frac{3 \cdot 2}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3} \text{ km/h} \]

Now, using the vertical component:
\[ V_r \cdot \frac{1}{2} = V_{vertical} \]
\[ V_{vertical} = \frac{2\sqrt{3}}{2} = \sqrt{3} \text{ km/h} \]

### Step 7: Calculate the magnitude of the rain's velocity
The actual speed of the rain can be calculated using Pythagoras theorem:
\[ V_r = \sqrt{(2\sqrt{3})^2 + (\sqrt{3})^2} \]
\[ = \sqrt{(4 \cdot 3) + 3} = \sqrt{12 + 3} = \sqrt{15} \text{ km/h} \]

### Final Answer
The speed of the rain drops is \( \sqrt{15} \) km/h, which is approximately \( 3.87 \) km/h.