An aeroplane A is flying at a level height towards north with a velocity of 800 km./hr as observed from the ground. Its pilot observes that another aeroplane B at the same height appears 10 him to be flying towards west with a velocity of 600 km/hr., The direction and magnitude of the velocity of the aeroplane B as observed from the ground are

Options

A. `500 , tan^(-1)"" (3)/(4)`
B. `1100 , tan ^(-1) "" (4)/(3)`
C. `100 , tan ^(-1) "" (4)/(3)`
D. none

Answer

To find the direction and magnitude of the velocity of aeroplane B as observed from the ground, we can use vector addition.

### Step-by-Step Solution:

1. **Identify the velocities**:
- The velocity of aeroplane A (V_A) is 800 km/hr towards the north.
- The velocity of aeroplane B (V_B) appears to the pilot of A to be 600 km/hr towards the west.

2. **Set up a coordinate system**:
- Let north be the positive y-direction and east be the positive x-direction.
- Therefore, the velocity of A can be represented as:
\[ V_A = 0 \hat{i} + 800 \hat{j} \text{ km/hr} \]
- The velocity of B as observed from A can be represented as:
\[ V_{B/A} = -600 \hat{i} + 0 \hat{j} \text{ km/hr} \]

3. **Use the relative velocity formula**:
- The velocity of B as observed from the ground (V_B) can be found using the formula:
\[ V_B = V_A + V_{B/A} \]
- Substituting the values:
\[ V_B = (0 \hat{i} + 800 \hat{j}) + (-600 \hat{i} + 0 \hat{j}) \]
\[ V_B = -600 \hat{i} + 800 \hat{j} \text{ km/hr} \]

4. **Calculate the magnitude of V_B**:
- The magnitude of the velocity vector can be calculated using the Pythagorean theorem:
\[ |V_B| = \sqrt{(-600)^2 + (800)^2} \]
\[ |V_B| = \sqrt{360000 + 640000} = \sqrt{1000000} = 1000 \text{ km/hr} \]

5. **Determine the direction of V_B**:
- The direction can be found using the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{800}{600} \]
\[ \theta = \tan^{-1}\left(\frac{800}{600}\right) = \tan^{-1}\left(\frac{4}{3}\right) \]
- Calculating this gives:
\[ \theta \approx 53.13^\circ \]
- Since V_B has a negative x-component and a positive y-component, it is in the second quadrant. Thus, the direction of V_B is:
\[ 180^\circ - 53.13^\circ = 126.87^\circ \text{ from the positive x-axis (east)} \]

### Final Answer:
- The magnitude of the velocity of aeroplane B as observed from the ground is 1000 km/hr.
- The direction is approximately 126.87° from the positive x-axis (east), or equivalently, 53.13° north of west.