A body released from a great height, falls freely towards the earth, Anoth . Another body is released from the same height ezectly one second later. Then the separation between two bodies, two seconds after the release fo second body is.

Answer

To solve the problem, we need to determine the separation between two bodies released from the same height, with one body released one second after the other. We will calculate the distance each body has fallen after a certain time and then find the separation between them.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let Body A be the first body released from a height \( h \) at time \( t = 0 \).
- Let Body B be the second body released from the same height \( h \) at time \( t = 1 \) second.
- We need to find the separation between the two bodies 2 seconds after Body B is released, which means we will calculate their positions at \( t = 3 \) seconds (1 second for Body A and 2 seconds for Body B).

2. **Calculate the Distance Fallen by Body A**:
- The time of fall for Body A when we are observing at \( t = 3 \) seconds is 3 seconds.
- Using the equation of motion:
\[ S_A = ut + \frac{1}{2} g t^2 \]
where \( u = 0 \) (initial velocity), \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity), and \( t = 3 \) seconds.
- Plugging in the values:
\[ S_A = 0 \cdot 3 + \frac{1}{2} \cdot 10 \cdot (3^2) = 0 + \frac{1}{2} \cdot 10 \cdot 9 = 45 \, \text{meters} \]

3. **Calculate the Distance Fallen by Body B**:
- The time of fall for Body B when we are observing at \( t = 3 \) seconds is 2 seconds.
- Using the same equation of motion:
\[ S_B = ut + \frac{1}{2} g t^2 \]
where \( u = 0 \), \( g = 10 \, \text{m/s}^2 \), and \( t = 2 \) seconds.
- Plugging in the values:
\[ S_B = 0 \cdot 2 + \frac{1}{2} \cdot 10 \cdot (2^2) = 0 + \frac{1}{2} \cdot 10 \cdot 4 = 20 \, \text{meters} \]

4. **Calculate the Separation Between the Two Bodies**:
- The separation \( d \) between the two bodies after 2 seconds of Body B's release is given by:
\[ d = S_A - S_B \]
- Substituting the distances we calculated:
\[ d = 45 \, \text{meters} - 20 \, \text{meters} = 25 \, \text{meters} \]

5. **Final Answer**:
- The separation between the two bodies 2 seconds after the release of the second body is **25 meters**.