A packet is dropped from a flight ascending with a velocity at height of 10km. If the packet takes 100s to reach the ground, then the velocity of the flight is (take `g= 10 ms^(-2)`)

Options

A. `100 ms^(-1)`
B. `300 ms^(-1)`
C. `200ms^(-1)`
D. `400 ms^(-1)`

Explanation

To solve the problem, we need to determine the velocity of the flight from which the packet is dropped. We know the following:

- Height (h) = 10 km = 10,000 m
- Time taken to reach the ground (t) = 100 s
- Acceleration due to gravity (g) = 10 m/s²

### Step-by-Step Solution:

1. **Determine the distance fallen by the packet**:
The packet is dropped from a height of 10 km. Therefore, the distance fallen by the packet is:
\[ h = 10,000 \text{ m} \]

2. **Use the equation of motion**:
The packet is in free fall, and we can use the second equation of motion to find the initial velocity of the packet (which is equal to the velocity of the flight):
\[ h = u t + \frac{1}{2} g t^2 \]
where:
- \( h \) = height (10,000 m)
- \( u \) = initial velocity of the packet (which is the same as the velocity of the flight)
- \( t \) = time taken (100 s)
- \( g \) = acceleration due to gravity (10 m/s²)

3. **Substitute the known values into the equation**:
\[ 10,000 = u \cdot 100 + \frac{1}{2} \cdot 10 \cdot (100)^2 \]

4. **Calculate the second term**:
\[ \frac{1}{2} \cdot 10 \cdot (100)^2 = 5 \cdot 10,000 = 50,000 \]

5. **Rearranging the equation**:
Now we can substitute this back into the equation:
\[ 10,000 = 100u + 50,000 \]

6. **Isolate \( u \)**:
\[ 100u = 10,000 - 50,000 \]
\[ 100u = -40,000 \]
\[ u = \frac{-40,000}{100} = -400 \text{ m/s} \]

7. **Interpret the result**:
The negative sign indicates that the velocity of the flight is directed upwards while the packet is falling downwards. Therefore, the magnitude of the velocity of the flight is:
\[ |u| = 400 \text{ m/s} \]

### Final Answer:
The velocity of the flight is **400 m/s**.