A person is walking to the east at 2 km/hr and the rain drops appear to him dropping vertically downwards at `2sqrt(3)km//hr`. Find the actual velocity of rain.

Answer

To solve the problem of finding the actual velocity of rain given the velocity of a person walking and the apparent velocity of the rain, we can follow these steps:

### Step 1: Define the velocities
- Let the velocity of the person walking east be \( \vec{v_p} = 2 \, \text{km/hr} \, \hat{i} \).
- The apparent velocity of the rain with respect to the person is given as \( \vec{v_{r/p}} = -2\sqrt{3} \, \text{km/hr} \, \hat{j} \) (since it appears to fall vertically downwards).

### Step 2: Use the relative velocity concept
The actual velocity of the rain with respect to the ground can be expressed using the relative velocity formula:
\[ \vec{v_r} = \vec{v_{r/p}} + \vec{v_p} \]
where \( \vec{v_r} \) is the actual velocity of the rain, \( \vec{v_{r/p}} \) is the velocity of rain with respect to the person, and \( \vec{v_p} \) is the velocity of the person.

### Step 3: Substitute the values
Substituting the known values into the equation:
\[ \vec{v_r} = (-2\sqrt{3} \, \hat{j}) + (2 \, \hat{i}) \]
This gives:
\[ \vec{v_r} = 2 \, \hat{i} - 2\sqrt{3} \, \hat{j} \]

### Step 4: Calculate the magnitude of the actual velocity of the rain
To find the magnitude of the actual velocity of the rain, we use the Pythagorean theorem:
\[ |\vec{v_r}| = \sqrt{(2)^2 + (-2\sqrt{3})^2} \]
Calculating the squares:
\[ |\vec{v_r}| = \sqrt{4 + 4 \cdot 3} = \sqrt{4 + 12} = \sqrt{16} = 4 \, \text{km/hr} \]

### Step 5: State the direction of the actual velocity of the rain
The direction of the actual velocity can be described by the vector \( \vec{v_r} = 2 \, \hat{i} - 2\sqrt{3} \, \hat{j} \), which indicates that the rain is falling at an angle with respect to the horizontal (east direction).

### Final Answer
The actual velocity of the rain is \( 4 \, \text{km/hr} \) in the direction of \( 2 \, \hat{i} - 2\sqrt{3} \, \hat{j} \).

---