An aeroplane pilot wishes to fly due west A wind of `100 km h^-1` is blowing towards south.
(a) If the speed of the plane (its speed in still air) is `300 km h^-1` in which direction should the pilot head ?
(b) What is the speed of the plane with respect tom ground ? Illustrate with a vector diagram.

Options

A. `sin^(-1)(2/3)` south of west
B. `30^(@)` East of north
C. `sin^(-1)(1/3)` north of west
D. none of these

Explanation

To solve the problem step by step, we will break it down into two parts as given in the question.

### Part (a): Direction in which the pilot should head

1. **Identify the vectors involved**:
- The velocity of the plane in still air (V_p) = 300 km/h (heading due west).
- The velocity of the wind (V_w) = 100 km/h (blowing towards south).

2. **Draw the vector diagram**:
- Draw a horizontal line pointing to the left to represent the westward direction (V_p).
- Draw a vertical line pointing downwards to represent the southward wind (V_w).

3. **Determine the resultant vector**:
- The pilot needs to adjust the heading to account for the wind. The effective velocity of the plane with respect to the ground (V_pg) can be represented as:
\[ V_{pg} = V_p + (-V_w) \]
- This means we need to find the angle θ such that the resultant vector points directly west.

4. **Use trigonometry to find the angle θ**:
- In the right triangle formed by V_p and V_w, we can use the sine function:
\[ \sin(\theta) = \frac{V_w}{V_p} = \frac{100}{300} = \frac{1}{3} \]
- Therefore,
\[ \theta = \sin^{-1}\left(\frac{1}{3}\right) \]
- This angle θ is north of west.

5. **Conclusion for part (a)**:
- The pilot should head at an angle of \( \sin^{-1}\left(\frac{1}{3}\right) \) north of west.

### Part (b): Speed of the plane with respect to the ground

1. **Calculate the speed of the plane with respect to the ground**:
- The speed of the plane with respect to the ground can be calculated using the cosine of the angle θ:
\[ V_{pg} = V_p \cdot \cos(\theta) \]
- We know:
\[ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \]

2. **Substituting the values**:
- Now substituting back into the equation for V_pg:
\[ V_{pg} = 300 \cdot \frac{2\sqrt{2}}{3} = 200\sqrt{2} \text{ km/h} \]

3. **Conclusion for part (b)**:
- The speed of the plane with respect to the ground is \( 200\sqrt{2} \) km/h.

### Summary of the solution:
- **Part (a)**: The pilot should head at an angle of \( \sin^{-1}\left(\frac{1}{3}\right) \) north of west.
- **Part (b)**: The speed of the plane with respect to the ground is \( 200\sqrt{2} \) km/h.