A man in a lift throws a ball vertically upwards with a velocity v and catches it (i) after `t_(1)` second when the lift is ascending with an upward acceleration of 'a' and (ii) after `t_(2)` second when the lift is descending downward with the same acceleration 'a'. THe velocity of the ball v is

Options

A. g `t_(1)t_(2)`
B. (1/2) g `t_(1)t_(2)`
C. 2 g `t_(1)t_(2)//t_(1) + t_(2)`
D. g `t_(1)t_(2)//t_(1) + t_(2)`

Explanation

To solve the problem, we need to analyze the motion of the ball in two different scenarios: when the lift is ascending and when it is descending.

### Step-by-Step Solution:

1. **Understanding the Scenario**:
- A man in a lift throws a ball vertically upwards with an initial velocity \( v \).
- The lift has an upward acceleration \( a \) in the first case and a downward acceleration \( a \) in the second case.
- The ball is caught after \( t_1 \) seconds in the first case and \( t_2 \) seconds in the second case.

2. **Effective Acceleration in the Lift**:
- When the lift is ascending with acceleration \( a \):
- The effective acceleration acting on the ball is \( g + a \) (downward).
- When the lift is descending with acceleration \( a \):
- The effective acceleration acting on the ball is \( g - a \) (upward).

3. **Using the Equations of Motion**:
- For the first case (lift ascending):
- The equation of motion can be expressed as:
\[ v - (-v) = (g + a) t_1 \]
Simplifying gives:
\[ 2v = (g + a) t_1 \quad \text{(1)} \]

- For the second case (lift descending):
- The equation of motion can be expressed as:
\[ v - (-v) = (g - a) t_2 \]
Simplifying gives:
\[ 2v = (g - a) t_2 \quad \text{(2)} \]

4. **Setting Up the Equations**:
- From equations (1) and (2), we have:
\[ 2v = (g + a) t_1 \quad \text{(1)} \]
\[ 2v = (g - a) t_2 \quad \text{(2)} \]

5. **Eliminating \( a \)**:
- We can add both equations to eliminate \( a \):
\[ (g + a) t_1 + (g - a) t_2 = 4v \]
This simplifies to:
\[ g(t_1 + t_2) + a(t_1 - t_2) = 4v \]
- Now, we can solve for \( a \):
\[ a(t_1 - t_2) = 4v - g(t_1 + t_2) \]
- Rearranging gives:
\[ a = \frac{4v - g(t_1 + t_2)}{t_1 - t_2} \]

6. **Finding the Value of \( v \)**:
- Now, we can substitute \( a \) back into one of the original equations to solve for \( v \). Let's use equation (1):
\[ 2v = (g + a) t_1 \]
- Substitute \( a \):
\[ 2v = \left(g + \frac{4v - g(t_1 + t_2)}{t_1 - t_2}\right) t_1 \]
- Rearranging and simplifying will yield:
\[ v = \frac{g t_1 t_2}{t_1 + t_2} \]

### Final Result:
Thus, the velocity of the ball \( v \) is given by:
\[ v = \frac{g t_1 t_2}{t_1 + t_2} \]