An aeroplane takes off from Mumbai to Delhi with velocity 50 kph at an angle `30^(@)` north of east. Wind is blowing at 25 kph from north to south. What is the resultant displacement of aeroplane in 2 hrs. (in km)

Explanation

To solve the problem of the aeroplane's resultant displacement, we can follow these steps:

### Step 1: Identify the Components of the Aeroplane's Velocity
The aeroplane is flying at a velocity of 50 kph at an angle of 30 degrees north of east. We can break this velocity into its x (east) and y (north) components using trigonometric functions.

- **East Component (Vx)**:
\[ V_x = V \cdot \cos(\theta) = 50 \cdot \cos(30^\circ) = 50 \cdot \frac{\sqrt{3}}{2} = 25\sqrt{3} \text{ kph} \]

- **North Component (Vy)**:
\[ V_y = V \cdot \sin(\theta) = 50 \cdot \sin(30^\circ) = 50 \cdot \frac{1}{2} = 25 \text{ kph} \]

### Step 2: Account for the Wind's Effect
The wind is blowing from north to south at 25 kph. This wind will affect the northward component of the aeroplane's velocity.

- The wind's velocity is in the negative y-direction (south), so we subtract it from the north component of the aeroplane's velocity:
\[ V_{y, \text{resultant}} = V_y - \text{wind speed} = 25 - 25 = 0 \text{ kph} \]

### Step 3: Calculate the Resultant Velocity
Since the northward component of the velocity cancels out due to the wind, the resultant velocity of the aeroplane is only in the east direction.

- Thus, the resultant velocity (Vr) is:
\[ V_r = V_x = 25\sqrt{3} \text{ kph} \]

### Step 4: Calculate the Displacement Over Time
To find the displacement over a time period of 2 hours, we use the formula:
\[ \text{Displacement} = \text{Velocity} \times \text{Time} \]
Substituting the values:
\[ \text{Displacement} = V_r \times t = 25\sqrt{3} \times 2 = 50\sqrt{3} \text{ km} \]

### Step 5: Convert to Decimal Form
To express the displacement in decimal form, we calculate \(50\sqrt{3}\):
\[ \sqrt{3} \approx 1.732 \implies 50\sqrt{3} \approx 50 \times 1.732 \approx 86.6 \text{ km} \]

### Final Answer
The resultant displacement of the aeroplane in 2 hours is approximately **86.6 km**.

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