Two bodies begin to fall freely from the same height but the second falls T second after the first. The time (after which the first body begins to fall) when the distance between the bodies equals L is

Options

A. `(1)/(2) T`
B. ` (T)/(2) +(L) /( gT) `
C. ` (L)/(gT) `
D. ` T+( 2L)/(gT) `

Explanation

To solve the problem, we need to analyze the motion of two bodies falling freely under gravity. Let's denote the first body as Body 1 and the second body as Body 2.

### Step-by-Step Solution:

1. **Define Variables**:
- Let \( t \) be the time after which the distance between the two bodies equals \( L \).
- Let \( T \) be the time delay after which Body 2 starts falling.
- The acceleration due to gravity is \( g \).

2. **Distance Covered by Body 1**:
- Body 1 falls for time \( t \).
- Using the equation of motion, the distance \( S_1 \) covered by Body 1 is given by:
\[ S_1 = \frac{1}{2} g t^2 \]

3. **Distance Covered by Body 2**:
- Body 2 starts falling \( T \) seconds after Body 1, so it falls for \( t - T \) seconds.
- The distance \( S_2 \) covered by Body 2 is:
\[ S_2 = \frac{1}{2} g (t - T)^2 \]

4. **Distance Between the Two Bodies**:
- The distance between the two bodies when it equals \( L \) can be expressed as:
\[ S_1 - S_2 = L \]
- Substituting the expressions for \( S_1 \) and \( S_2 \):
\[ \frac{1}{2} g t^2 - \frac{1}{2} g (t - T)^2 = L \]

5. **Simplifying the Equation**:
- Expanding \( (t - T)^2 \):
\[ (t - T)^2 = t^2 - 2tT + T^2 \]
- Thus, we have:
\[ \frac{1}{2} g t^2 - \frac{1}{2} g (t^2 - 2tT + T^2) = L \]
- Simplifying this gives:
\[ \frac{1}{2} g (2tT - T^2) = L \]
- Rearranging leads to:
\[ g t T - \frac{1}{2} g T^2 = L \]

6. **Solving for Time \( t \)**:
- Rearranging the equation:
\[ g t T = L + \frac{1}{2} g T^2 \]
- Thus, solving for \( t \):
\[ t = \frac{L + \frac{1}{2} g T^2}{g T} \]

### Final Expression:
The time \( t \) after which the distance between the two bodies equals \( L \) is:
\[ t = \frac{L}{g T} + \frac{T}{2} \]