A man standing on a road has to hold his umbrella at `30^(@)` with the vertical to keep the rain away. He throws the umbrella and starts running at 10 kmph. He find that rain drops are hitting his head vertically. The speed of rain drops relative to the ground is

Options

A. 10 kmph
B. 20 kmph
C. `10 sqrt(3)` kmph
D. `20 sqrt(3)` kmph

Answer

To solve the problem, we need to analyze the motion of the man and the rain drops in relation to each other. Here’s a step-by-step solution:

### Step 1: Understand the Situation
The man initially holds his umbrella at an angle of \(30^\circ\) with the vertical to keep the rain away. This means that the rain has a horizontal component of velocity that he is countering with the umbrella.

### Step 2: Set Up the Diagram
Draw a right triangle where:
- The vertical side represents the vertical component of the rain's velocity (\(V_{R_y}\)).
- The horizontal side represents the horizontal component of the rain's velocity (\(V_{R_x}\)).
- The angle with the vertical is \(30^\circ\).

### Step 3: Relate the Components of Velocity
From the triangle:
- The tangent of the angle gives us the relationship between the horizontal and vertical components:
\[ \tan(30^\circ) = \frac{V_{R_x}}{V_{R_y}} \]

Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), we can write:
\[ \frac{1}{\sqrt{3}} = \frac{V_{R_x}}{V_{R_y}} \]
This implies:
\[ V_{R_x} = \frac{V_{R_y}}{\sqrt{3}} \]

### Step 4: Consider the Man's Motion
When the man starts running at \(10 \text{ km/h}\), he finds that the rain is hitting him vertically. This means that the horizontal component of the rain's velocity (\(V_{R_x}\)) must equal the speed at which the man is running:
\[ V_{R_x} = 10 \text{ km/h} \]

### Step 5: Substitute to Find \(V_{R_y}\)
Using the relationship we found earlier:
\[ 10 = \frac{V_{R_y}}{\sqrt{3}} \]
Multiplying both sides by \(\sqrt{3}\):
\[ V_{R_y} = 10\sqrt{3} \text{ km/h} \]

### Step 6: Find the Resultant Velocity of Rain
Now, we can find the speed of the rain relative to the ground. The speed of the rain (\(V_R\)) can be found using the Pythagorean theorem:
\[ V_R = \sqrt{V_{R_x}^2 + V_{R_y}^2} \]
Substituting the values:
\[ V_R = \sqrt{(10)^2 + (10\sqrt{3})^2} \]
\[ = \sqrt{100 + 300} = \sqrt{400} = 20 \text{ km/h} \]

### Conclusion
Thus, the speed of the rain drops relative to the ground is \(20 \text{ km/h}\).