An aeroplane is flying in vertical plane at an angle of `30^(@)` with the horizontal (north) and wind is is blowing from west.A package is dropped from an aeroplane. The velocity of the wind if package hits a kite flying in the space with a position vector `vec(R) = (400 sqrt(3) hat(i) + 80 hat(j) + 200 hat(k))`m with respect to the point of dropping. (Here `hat(i)` and `hat(j)` are the unit vectors along north and vertically up respectively and `hat(k)` be the unit vector due east. Assume that the bag is light enough to get carried away by the wind)

Options

A. 50 m/sec
B. 25 m/sec
C. 20 m/sec
D. 10 m/sec

Explanation

To solve the problem step by step, we will analyze the motion of the package dropped from the airplane and equate it to the position of the kite to find the wind velocity.

### Step 1: Understand the motion of the package
The airplane is flying at an angle of \(30^\circ\) with the horizontal (north). The velocity of the package dropped from the airplane can be broken down into components:
- Horizontal (north) component: \(u \cos(30^\circ)\)
- Vertical (upward) component: \(u \sin(30^\circ)\)
- Eastward component due to wind: \(V_w\)

### Step 2: Write the position vector of the package
The position vector of the package at time \(t\) after being dropped can be expressed as:
\[ \vec{R}_{package} = (u \cos(30^\circ) t) \hat{i} + \left(u \sin(30^\circ) t - \frac{1}{2} g t^2\right) \hat{j} + (V_w t) \hat{k} \]

### Step 3: Substitute the values of \( \cos(30^\circ} \) and \( \sin(30^\circ} \)
Using the values:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2}, \quad \sin(30^\circ) = \frac{1}{2} \]
The position vector becomes:
\[ \vec{R}_{package} = \left(u \frac{\sqrt{3}}{2} t\right) \hat{i} + \left(u \frac{1}{2} t - \frac{1}{2} g t^2\right) \hat{j} + (V_w t) \hat{k} \]

### Step 4: Set up the position vector of the kite
The position vector of the kite is given as:
\[ \vec{R}_{kite} = (400 \sqrt{3}) \hat{i} + 80 \hat{j} + 200 \hat{k} \]

### Step 5: Equate the position vectors
For the package to hit the kite, their position vectors must be equal:
\[ \left(u \frac{\sqrt{3}}{2} t\right) = 400 \sqrt{3} \quad \text{(1)} \]
\[ \left(u \frac{1}{2} t - \frac{1}{2} g t^2\right) = 80 \quad \text{(2)} \]
\[ (V_w t) = 200 \quad \text{(3)} \]

### Step 6: Solve equation (1) for \(u\)
From equation (1):
\[ u \frac{\sqrt{3}}{2} t = 400 \sqrt{3} \]
\[ u t = 800 \quad \text{(4)} \]

### Step 7: Solve equation (2) for \(t\)
Substituting \(u\) from equation (4) into equation (2):
\[ \frac{800}{2} - \frac{1}{2} g t^2 = 80 \]
\[ 400 - 5t^2 = 80 \]
\[ 5t^2 = 320 \]
\[ t^2 = 64 \implies t = 8 \text{ seconds} \]

### Step 8: Solve equation (3) for \(V_w\)
Substituting \(t\) into equation (3):
\[ V_w t = 200 \]
\[ V_w \cdot 8 = 200 \]
\[ V_w = \frac{200}{8} = 25 \text{ m/s} \]

### Final Answer
The velocity of the wind is \(V_w = 25 \text{ m/s}\).

---