A hot-air balloon is ascending at a rate of 14 m/s at a height of 98 m above the ground when a packet is dropped from it. (a) With what speed does the packet reach the ground, and (b) how much time does the fall take ?

Answer

To solve the problem step by step, let's break it down into two parts: (a) finding the speed of the packet when it reaches the ground, and (b) calculating the time it takes for the packet to fall.

### Given Data:
- Initial height (h) = 98 m
- Initial velocity of the packet (u) = 14 m/s (upward)
- Acceleration due to gravity (g) = 9.81 m/s² (downward)

### Part (a): Finding the speed of the packet when it reaches the ground

1. **Identify the motion parameters**:
- The initial velocity (u) of the packet is 14 m/s upward.
- The downward displacement (s) is -98 m (since it is falling down).
- The acceleration (a) due to gravity is -9.81 m/s² (downward).

2. **Use the third equation of motion**:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity (what we need to find)
- \( u \) = initial velocity = 14 m/s
- \( a \) = acceleration = -9.81 m/s²
- \( s \) = displacement = -98 m

3. **Substituting the values**:
\[ v^2 = (14)^2 + 2 \times (-9.81) \times (-98) \]
\[ v^2 = 196 + 2 \times 9.81 \times 98 \]
\[ v^2 = 196 + 1923.96 \]
\[ v^2 = 2120.96 \]

4. **Calculating \( v \)**:
\[ v = \sqrt{2120.96} \approx 46.0 \text{ m/s} \]
Since the packet is falling downwards, we take the negative value:
\[ v \approx -46.0 \text{ m/s} \]

### Part (b): Finding the time taken to fall

1. **Use the first equation of motion**:
\[ v = u + at \]
Rearranging for time \( t \):
\[ t = \frac{v - u}{a} \]

2. **Substituting the known values**:
\[ t = \frac{-46 - 14}{-9.81} \]
\[ t = \frac{-60}{-9.81} \approx 6.12 \text{ seconds} \]

### Final Answers:
(a) The speed of the packet when it reaches the ground is approximately **46.0 m/s** (downward).

(b) The time taken for the packet to fall is approximately **6.12 seconds**.

---