A food package was dropped from an aircraft flying horizontally. 6 s before it hit the ground, it was at a height of `780 m`, and had travelled a distance of 1 km horizontally. Find the speed and the altitude of the aircraft.

Explanation

To solve the problem, we need to find the speed of the aircraft and the altitude from which the food package was dropped. We know that the package was dropped from a height of 780 m and it traveled 1 km horizontally in 6 seconds before hitting the ground.

### Step 1: Find the speed of the aircraft

1. **Given Data**:
- Horizontal distance traveled by the package (s) = 1 km = 1000 m
- Time taken (t) = 6 s

2. **Using the formula for horizontal motion**:
\[ s = vt \]
Here, \(v\) is the speed of the aircraft.

3. **Rearranging the formula to find \(v\)**:
\[ v = \frac{s}{t} = \frac{1000 \text{ m}}{6 \text{ s}} = \frac{1000}{6} \text{ m/s} \approx 166.67 \text{ m/s} \]

### Step 2: Find the altitude of the aircraft

1. **Given Data**:
- Height of the package at the moment of dropping (h) = 780 m
- Time taken to hit the ground = 6 s

2. **Using the vertical motion equations**:
The vertical displacement can be calculated using the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \(s\) = vertical displacement = -780 m (downward)
- \(u\) = initial vertical velocity = 0 (since it is dropped)
- \(a\) = acceleration due to gravity = -10 m/s² (downward)
- \(t\) = time = 6 s

3. **Substituting the values into the equation**:
\[ -780 = 0 \cdot 6 + \frac{1}{2} \cdot (-10) \cdot (6^2) \]
\[ -780 = -5 \cdot 36 \]
\[ -780 = -180 \]
This confirms the vertical displacement calculation.

4. **Finding the total height from which the package was dropped**:
The total height (H) of the aircraft can be calculated as:
\[ H = h + \text{displacement} \]
\[ H = 780 + 780 = 1560 \text{ m} \]

### Final Answers:
- Speed of the aircraft: \(166.67 \text{ m/s}\)
- Altitude of the aircraft: \(1560 \text{ m}\)