An airplane is observed by two observers traveling at `60 km h^-1` in two vehicles moving in opposite directions on a straight road. To an observer in one vehicle, the plane appears to cross the road track at right angles while to the other appears to be `45^@`. At what angle does the plane actually cross the road track and what is its speed relative to ground ?

Explanation

To solve the problem, we need to find the angle at which the airplane crosses the road track and its speed relative to the ground. We have two observers moving in opposite directions at a speed of 60 km/h, and the airplane appears to cross the road at different angles to each observer.

### Step-by-Step Solution:

1. **Define the Velocity Components:**
- Let the velocity of the airplane be represented as \( V = V_x \hat{i} + V_y \hat{j} \).
- The first observer (Observer 1) is moving in the positive y-direction with a velocity of \( 60 \hat{j} \) km/h.
- The second observer (Observer 2) is moving in the negative y-direction with a velocity of \( -60 \hat{j} \) km/h.

2. **Relative Velocity for Observer 1:**
- The relative velocity of the airplane with respect to Observer 1 is:
\[ V_{1} = V - 60 \hat{j} = V_x \hat{i} + (V_y - 60) \hat{j} \]
- Since the airplane appears to cross the road at a right angle to the track for Observer 1, the y-component of the relative velocity must be zero:
\[ V_y - 60 = 0 \implies V_y = 60 \text{ km/h} \]

3. **Relative Velocity for Observer 2:**
- The relative velocity of the airplane with respect to Observer 2 is:
\[ V_{2} = V + 60 \hat{j} = V_x \hat{i} + (V_y + 60) \hat{j} \]
- For Observer 2, the airplane appears to cross the road at a \( 45^\circ \) angle. Therefore, the tangent of the angle must be equal to 1 (since \( \tan(45^\circ) = 1 \)):
\[ \tan(45^\circ) = \frac{V_y + 60}{V_x} = 1 \]
- This gives us the equation:
\[ V_y + 60 = V_x \implies V_x = V_y + 60 \]

4. **Substituting for \( V_y \):**
- We already found \( V_y = 60 \) km/h. Substituting this into the equation for \( V_x \):
\[ V_x = 60 + 60 = 120 \text{ km/h} \]

5. **Calculating the Speed of the Airplane:**
- The speed of the airplane relative to the ground can be calculated using the Pythagorean theorem:
\[ V = \sqrt{V_x^2 + V_y^2} = \sqrt{(120)^2 + (60)^2} \]
- Calculating this gives:
\[ V = \sqrt{14400 + 3600} = \sqrt{18000} = 60\sqrt{5} \text{ km/h} \]

6. **Calculating the Angle of the Airplane's Path:**
- The angle \( \phi \) at which the airplane crosses the road can be found using:
\[ \tan(\phi) = \frac{V_y}{V_x} = \frac{60}{120} = \frac{1}{2} \]
- Therefore, the angle \( \phi \) is:
\[ \phi = \tan^{-1}\left(\frac{1}{2}\right) \]

### Final Answers:
- The angle at which the airplane actually crosses the road track is \( \tan^{-1}\left(\frac{1}{2}\right) \).
- The speed of the airplane relative to the ground is \( 60\sqrt{5} \) km/h.