Two bodies begin to fall freely from the same height. The second one begins to fall `tau` s after the first. The time after which the 2st body begins to fall, the distance between the bodies equals to l is

Options

A. `(l)/(g tau)-(tau)/(2)`
B. `(g tau)/(l)+tau`
C. `(tau)/(lg)+(2)/(tau)`
D. `(g)/(l tau)+(tau)/(2)`

Explanation

To solve the problem of two bodies falling freely from the same height, where the second body starts falling `tau` seconds after the first, we need to determine the time `t` after which the distance between the two bodies equals `l`.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- Both bodies are in free fall, meaning they are subjected to the acceleration due to gravity, `g`.
- The first body starts falling at `t = 0` and the second body starts falling at `t = tau`.

2. **Distance Covered by Each Body**:
- The distance covered by the first body after `t + tau` seconds (since it has been falling for `t + tau` seconds) is given by the equation:
\[ S_1 = \frac{1}{2} g (t + \tau)^2 \]
- The distance covered by the second body after `t` seconds (since it has been falling for `t` seconds) is:
\[ S_2 = \frac{1}{2} g t^2 \]

3. **Setting Up the Equation**:
- According to the problem, the distance between the two bodies when the second body has been falling for `t` seconds is equal to `l`. Therefore, we can set up the equation:
\[ S_1 - S_2 = l \]
- Substituting the expressions for `S_1` and `S_2`:
\[ \frac{1}{2} g (t + \tau)^2 - \frac{1}{2} g t^2 = l \]

4. **Simplifying the Equation**:
- Factor out \(\frac{1}{2} g\):
\[ \frac{1}{2} g \left( (t + \tau)^2 - t^2 \right) = l \]
- Expanding the left side:
\[ (t + \tau)^2 - t^2 = t^2 + 2t\tau + \tau^2 - t^2 = 2t\tau + \tau^2 \]
- Thus, the equation becomes:
\[ \frac{1}{2} g (2t\tau + \tau^2) = l \]

5. **Rearranging to Find `t`**:
- Multiply both sides by 2:
\[ g (2t\tau + \tau^2) = 2l \]
- Rearranging gives:
\[ 2t\tau + \tau^2 = \frac{2l}{g} \]
- Solving for `t`:
\[ 2t\tau = \frac{2l}{g} - \tau^2 \]
\[ t = \frac{1}{2\tau} \left( \frac{2l}{g} - \tau^2 \right) \]

6. **Final Expression for `t`**:
- Simplifying further:
\[ t = \frac{l}{g\tau} - \frac{\tau}{2} \]

### Final Answer:
The time `t` after which the distance between the two bodies equals `l` is given by:
\[ t = \frac{l}{g\tau} - \frac{\tau}{2} \]