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

Options

A. 4.9 m
B. 9.8 m
C. 19.6 m
D. 24.5 m

Explanation

To solve the problem, we need to determine the separation between two bodies falling freely under gravity after a specific time. Let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Body B1 is released at \( t = 0 \) seconds.
- Body B2 is released at \( t = 1 \) second.
- We need to find the separation between the two bodies 2 seconds after B2 is released, which is at \( t = 3 \) seconds.

2. **Distance Traveled by Body B1**:
- We need to calculate the distance traveled by B1 in 3 seconds.
- The formula for distance under constant acceleration (free fall) is:
\[ s = ut + \frac{1}{2} a t^2 \]
- Here, \( u = 0 \) (initial velocity), \( a = g \) (acceleration due to gravity), and \( t = 3 \) seconds.
- Thus, the distance \( s_1 \) traveled by B1 is:
\[ s_1 = 0 \cdot 3 + \frac{1}{2} g (3^2) = \frac{1}{2} g \cdot 9 = \frac{9}{2} g \]

3. **Distance Traveled by Body B2**:
- Now, we calculate the distance traveled by B2 in 2 seconds (from \( t = 1 \) second to \( t = 3 \) seconds).
- Using the same formula, for B2:
- Here, \( t = 2 \) seconds.
- Thus, the distance \( s_2 \) traveled by B2 is:
\[ s_2 = 0 \cdot 2 + \frac{1}{2} g (2^2) = \frac{1}{2} g \cdot 4 = 2g \]

4. **Finding the Separation**:
- The separation between the two bodies at \( t = 3 \) seconds is given by:
\[ \text{Separation} = s_1 - s_2 \]
- Substituting the values we found:
\[ \text{Separation} = \frac{9}{2} g - 2g \]
- To combine the terms, convert \( 2g \) into a fraction:
\[ 2g = \frac{4}{2} g \]
- Thus, the separation becomes:
\[ \text{Separation} = \frac{9}{2} g - \frac{4}{2} g = \frac{5}{2} g \]

5. **Calculating the Numerical Value**:
- Using \( g \approx 9.8 \, \text{m/s}^2 \):
\[ \text{Separation} = \frac{5}{2} \cdot 9.8 = \frac{49}{2} = 24.5 \, \text{meters} \]

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