A body is released from a great height and falls freely towards the earth. Exactly one sec later another body is released. What is the distance between the two bodies 2 sec after the release of the second body ?

Options

A. `15^(@)`
B. `30^(@)`
C. `90^(@)`
D. `45^(@)`

Explanation

To solve the problem, we need to calculate the positions of both bodies at a specific time and then find the distance between them.

### Step-by-Step Solution:

1. **Understand the scenario**:
- A body (Body 1) is released from a great height at \( t = 0 \) seconds.
- Another body (Body 2) is released 1 second later, at \( t = 1 \) second.
- We need to find the distance between the two bodies 2 seconds after the release of Body 2, which means at \( t = 3 \) seconds.

2. **Calculate the time of fall for each body**:
- For Body 1, it has been falling for 3 seconds by the time we are considering (from \( t = 0 \) to \( t = 3 \)).
- For Body 2, it has been falling for 2 seconds (from \( t = 1 \) to \( t = 3 \)).

3. **Use the second equation of motion**:
- The second equation of motion states that the distance fallen by an object under free fall is given by:
\[ h = \frac{1}{2} g t^2 \]
- Where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

4. **Calculate the distance fallen by Body 1**:
- For Body 1 at \( t_1 = 3 \) seconds:
\[ h_1 = \frac{1}{2} g (3)^2 = \frac{1}{2} g \cdot 9 = \frac{9g}{2} \]

5. **Calculate the distance fallen by Body 2**:
- For Body 2 at \( t_2 = 2 \) seconds:
\[ h_2 = \frac{1}{2} g (2)^2 = \frac{1}{2} g \cdot 4 = \frac{4g}{2} = 2g \]

6. **Find the distance between the two bodies**:
- The distance between the two bodies is given by:
\[ \text{Distance} = h_1 - h_2 = \frac{9g}{2} - 2g \]
- Converting \( 2g \) to a fraction with the same denominator:
\[ 2g = \frac{4g}{2} \]
- Now, substituting this in:
\[ \text{Distance} = \frac{9g}{2} - \frac{4g}{2} = \frac{5g}{2} \]

7. **Substituting the value of \( g \)**:
- Using \( g \approx 9.8 \, \text{m/s}^2 \):
\[ \text{Distance} = \frac{5 \times 9.8}{2} = \frac{49}{2} = 24.5 \, \text{meters} \]

### Final Answer:
The distance between the two bodies 2 seconds after the release of the second body is approximately **24.5 meters**.