An aeroplane is flying with the velocity of `V_(1)` =800 kmph relative to the air towards south. A wind with velocity of `V_(2) =15 ms^(-1)` is blowing from West to East. What is the velocity of the aeroplane with respect to the earth.

Options

A. `222.7 ms^(-1)`
B. `150 ms^(-1)`
C. `82 ms^(-1)`
D. `40 ms^(-1)`

Answer

To find the velocity of the aeroplane with respect to the Earth, we can follow these steps:

### Step 1: Convert the velocity of the aeroplane from km/h to m/s
The velocity of the aeroplane \( V_1 \) is given as 800 km/h. We need to convert this into meters per second (m/s).

\[ V_1 = 800 \, \text{km/h} = \frac{800 \times 1000 \, \text{m}}{3600 \, \text{s}} = \frac{800000}{3600} \approx 222.22 \, \text{m/s} \]

### Step 2: Define the velocity vectors
- The velocity of the aeroplane \( V_1 \) is directed south, which we can represent as:
\[ \vec{V_1} = -222.22 \, \hat{j} \, \text{m/s} \]
- The wind velocity \( V_2 \) is blowing from west to east at 15 m/s, which we can represent as:
\[ \vec{V_2} = 15 \, \hat{i} \, \text{m/s} \]

### Step 3: Use vector addition to find the velocity of the aeroplane with respect to the Earth
The velocity of the aeroplane with respect to the Earth \( \vec{V_{AE}} \) can be found using the equation:
\[ \vec{V_{AE}} = \vec{V_1} + \vec{V_2} \]
Substituting the values:
\[ \vec{V_{AE}} = (-222.22 \, \hat{j}) + (15 \, \hat{i}) = 15 \, \hat{i} - 222.22 \, \hat{j} \]

### Step 4: Calculate the magnitude of the resultant velocity vector
To find the magnitude of the velocity vector \( \vec{V_{AE}} \):
\[ |\vec{V_{AE}}| = \sqrt{(15)^2 + (-222.22)^2} \]
Calculating this:
\[ |\vec{V_{AE}}| = \sqrt{15^2 + (222.22)^2} = \sqrt{225 + 49333.7284} \approx \sqrt{49558.7284} \approx 222.6 \, \text{m/s} \]

### Step 5: Conclusion
The velocity of the aeroplane with respect to the Earth is approximately \( 222.6 \, \text{m/s} \).

---