A man in a lift ascending with an acceleration throws a ball vertically up with a velocity u and catches it after time `t_1`. Afterwards when the lift is descending with the same acceleration, the man again throws a ball vertically up with the same velocity and catches it after time `t_2`. The velocity of projection of ball is

Options

A. `("gt"_1 t_1)/((t_1 - t_2))`
B. `( g t_1 t_2)/((t_1 + t_2))`
C. `(g t_1 t_2)/(2(t_1 - t_2))`
D. `(g t_1 t_2)/(2(t_1 + t_2))`

Answer

To solve the problem, we need to analyze the two scenarios in which the man throws the ball from the lift, both when the lift is ascending and descending with the same acceleration.

### Step-by-Step Solution:

1. **Understanding the Scenario**:
- When the lift ascends with acceleration \( a \), and the man throws the ball upwards with an initial velocity \( u \), the effective acceleration acting on the ball will be \( g + a \) (where \( g \) is the acceleration due to gravity).
- When the lift descends with the same acceleration \( a \), the effective acceleration acting on the ball will be \( g - a \).

2. **Time of Flight in Ascending Lift**:
- The time of flight \( t_1 \) when the lift is ascending can be given by the formula:
\[ t_1 = \frac{2u}{g + a} \]
- Rearranging this gives us:
\[ u = \frac{t_1 (g + a)}{2} \quad \text{(Equation 1)} \]

3. **Time of Flight in Descending Lift**:
- The time of flight \( t_2 \) when the lift is descending can be given by the formula:
\[ t_2 = \frac{2u}{g - a} \]
- Rearranging this gives us:
\[ u = \frac{t_2 (g - a)}{2} \quad \text{(Equation 2)} \]

4. **Equating the Two Expressions for \( u \)**:
- From Equation 1 and Equation 2, we can set them equal to each other:
\[ \frac{t_1 (g + a)}{2} = \frac{t_2 (g - a)}{2} \]
- Multiplying through by 2 to eliminate the fraction:
\[ t_1 (g + a) = t_2 (g - a) \]

5. **Expanding and Rearranging**:
- Expanding both sides gives:
\[ t_1 g + t_1 a = t_2 g - t_2 a \]
- Rearranging terms to isolate \( a \):
\[ t_1 g + t_1 a + t_2 a = t_2 g \]
\[ a(t_1 + t_2) = t_2 g - t_1 g \]
\[ a(t_1 + t_2) = g(t_2 - t_1) \]
- Thus, we can solve for \( a \):
\[ a = \frac{g(t_2 - t_1)}{t_1 + t_2} \]

6. **Substituting Back to Find \( u \)**:
- Now substituting \( a \) back into either Equation 1 or Equation 2 to find \( u \). Using Equation 1:
\[ u = \frac{t_1 \left(g + \frac{g(t_2 - t_1)}{t_1 + t_2}\right)}{2} \]
- Simplifying this gives:
\[ u = \frac{t_1 \left(\frac{g(t_1 + t_2) + g(t_2 - t_1)}{t_1 + t_2}\right)}{2} \]
\[ u = \frac{t_1 \left(\frac{g(t_1 + t_2)}{t_1 + t_2}\right)}{2} \]
\[ u = \frac{t_1 t_2 g}{t_1 + t_2} \]

### Final Answer:
The velocity of projection of the ball is:
\[ u = \frac{g t_1 t_2}{t_1 + t_2} \]