A man in a lift ascending with an upward acceleration `a` throws a ball vertically upwards with a velocity `v` with respect to himself and catches it after `t_1` seconds. After wards when the lift is descending with the same acceleration `a` acting downwards the man again throws the ball vertically upwards with the same velocity with respect to him and catches it after `t_2` seconds?

Options

A. The acceleration of the ball w.r.t. ground is g when it is in air
B. the velocity v of the ball relative to the ligt is `(g(t_1+t_2))/(t_1+t_2)`
C. the acceleration `a` of the lift is `(g(t_2-t_1))/(t_1+t_2)`
D. The velocity `v` of the ball relative to the man is `(gt_1t_2)/((t_1+t_2))`

Explanation

To solve the problem step by step, we will analyze the motion of the ball in both scenarios: when the lift is ascending and when it is descending.

### Step 1: Analyze the first scenario (Lift ascending)
1. **Initial Conditions**:
- The lift is ascending with an upward acceleration \( a \).
- The man throws the ball vertically upwards with a velocity \( v \) with respect to himself.
- The time taken to catch the ball is \( t_1 \).

2. **Acceleration of the ball with respect to the ground**:
- When the ball is thrown, it experiences two accelerations: the gravitational acceleration \( g \) acting downwards and the upward acceleration \( a \) of the lift.
- The effective acceleration of the ball with respect to the ground is \( g - a \) (since both accelerations are in opposite directions).

3. **Using the equations of motion**:
- The ball will go up and then come down. The time to reach the maximum height is \( t_{up} \) and the time to come down is also \( t_{up} \).
- Therefore, the total time of flight \( t_1 = 2t_{up} \).

4. **Finding \( t_{up} \)**:
- At maximum height, the final velocity of the ball is 0. Using the equation of motion:
\[ 0 = v - (g - a)t_{up} \]
Rearranging gives:
\[ t_{up} = \frac{v}{g - a} \]
- Thus, the total time of flight is:
\[ t_1 = 2t_{up} = \frac{2v}{g - a} \]

### Step 2: Analyze the second scenario (Lift descending)
1. **Initial Conditions**:
- Now, the lift is descending with the same upward acceleration \( a \) acting downwards.
- The man again throws the ball vertically upwards with the same velocity \( v \) with respect to himself.
- The time taken to catch the ball is \( t_2 \).

2. **Acceleration of the ball with respect to the ground**:
- In this case, the effective acceleration of the ball with respect to the ground is \( g + a \) (since both accelerations are in the same direction).

3. **Using the equations of motion**:
- Similar to the first scenario, the total time of flight \( t_2 = 2t'_{up} \).

4. **Finding \( t'_{up} \)**:
- At maximum height, the final velocity of the ball is 0. Using the equation of motion:
\[ 0 = v - (g + a)t'_{up} \]
Rearranging gives:
\[ t'_{up} = \frac{v}{g + a} \]
- Thus, the total time of flight is:
\[ t_2 = 2t'_{up} = \frac{2v}{g + a} \]

### Step 3: Relating \( t_1 \) and \( t_2 \)
1. We have:
\[ t_1 = \frac{2v}{g - a} \quad \text{and} \quad t_2 = \frac{2v}{g + a} \]

2. To find the relationship between \( t_1 \) and \( t_2 \):
- Cross-multiplying gives:
\[ t_1 (g + a) = t_2 (g - a) \]
- Expanding this gives:
\[ t_1 g + t_1 a = t_2 g - t_2 a \]
- Rearranging leads to:
\[ t_1 g + t_2 a = t_2 g - t_1 a \]

### Step 4: Final Expressions
1. Rearranging the above equation gives:
\[ (t_1 + t_2) g = (t_1 + t_2) a \]
- This can be simplified to find the values of \( a \) and \( v \) in terms of \( t_1 \) and \( t_2 \).

### Summary of Results
- The acceleration \( a \) and the initial velocity \( v \) can be expressed as:
\[ a = \frac{g(t_2 - t_1)}{t_1 + t_2} \]
\[ v = \frac{g t_1 t_2}{t_1 + t_2} \]