A reackless drunk is playing with a gun in an airplane that is going directly east at `500 km h^(-1)`. The drunk shoots the gun straight up at the ceiling of the plane. The bullet leaves the gun at a speed of `1000 kmh^(-1)`. Relative to an observer on earth, what angle does the bullet make with the vertical ?

Answer

To solve the problem of determining the angle the bullet makes with the vertical as observed from the ground, we can break down the velocities involved and use trigonometry.

### Step-by-Step Solution:

1. **Identify the velocities**:
- The airplane is moving east at a velocity \( V_{plane} = 500 \, \text{km/h} \).
- The bullet is shot straight up with a velocity \( V_{bullet} = 1000 \, \text{km/h} \).

2. **Determine the components of the bullet's velocity**:
- The bullet has two components of velocity:
- Vertical component \( V_{y} = 1000 \, \text{km/h} \) (upward).
- Horizontal component \( V_{x} = 500 \, \text{km/h} \) (same direction as the plane).

3. **Calculate the resultant velocity of the bullet**:
- The resultant velocity \( V_{resultant} \) can be found using the Pythagorean theorem:
\[ V_{resultant} = \sqrt{V_{x}^2 + V_{y}^2} \]
- Substituting the values:
\[ V_{resultant} = \sqrt{(500)^2 + (1000)^2} = \sqrt{250000 + 1000000} = \sqrt{1250000} = 1118.03 \, \text{km/h} \]

4. **Determine the angle with respect to the vertical**:
- We can find the angle \( \theta \) using the tangent function:
\[ \tan(\theta) = \frac{V_{x}}{V_{y}} = \frac{500}{1000} = 0.5 \]
- To find the angle \( \theta \):
\[ \theta = \tan^{-1}(0.5) \]
- Calculating this gives:
\[ \theta \approx 26.57^\circ \]

5. **Conclusion**:
- The angle the bullet makes with the vertical is approximately \( 26.57^\circ \).

### Final Answer:
The bullet makes an angle of approximately \( 26.57^\circ \) with the vertical.