It is rainning vertically downwards. By how many times that a person should increase his velocity in order to change his umbrella position from an angle `30^(@)` with vertical to `60^(@)` vertical to protect himself from rain.

Options

A. 1
B. 2
C. 2.5
D. 3

Answer

To solve the problem, we need to analyze the situation using the concept of relative velocity. The person is trying to change the angle of his umbrella to protect himself from rain falling vertically downwards.

### Step-by-Step Solution:

1. **Understanding the Angles:**
- The umbrella is initially held at an angle of \(30^\circ\) with the vertical.
- The umbrella is then changed to an angle of \(60^\circ\) with the vertical.

2. **Setting Up the Velocity Components:**
- Let \(V_r\) be the velocity of the rain (which is vertical).
- Let \(V_m\) be the initial velocity of the man (horizontal).
- When the umbrella is at \(30^\circ\), the tangent of the angle gives us the relationship between the horizontal and vertical components:
\[ \tan(30^\circ) = \frac{V_m}{V_r} \]
- When the umbrella is at \(60^\circ\), the new horizontal velocity of the man will be \(nV_m\) (where \(n\) is the factor by which the velocity increases):
\[ \tan(60^\circ) = \frac{nV_m}{V_r} \]

3. **Using the Tangent Values:**
- We know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \quad \text{and} \quad \tan(60^\circ) = \sqrt{3} \]

4. **Setting Up the Equations:**
- From the first angle:
\[ \frac{1}{\sqrt{3}} = \frac{V_m}{V_r} \quad \text{(Equation 1)} \]
- From the second angle:
\[ \sqrt{3} = \frac{nV_m}{V_r} \quad \text{(Equation 2)} \]

5. **Dividing the Two Equations:**
- We divide Equation 1 by Equation 2:
\[ \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}} = \frac{\frac{V_m}{V_r}}{\frac{nV_m}{V_r}} \]
- This simplifies to:
\[ \frac{1}{3} = \frac{1}{n} \]

6. **Solving for \(n\):**
- Rearranging gives:
\[ n = 3 \]

### Final Answer:
The person should increase his velocity by a factor of **3** to change his umbrella position from \(30^\circ\) to \(60^\circ\) with the vertical.