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. **Given**: The lift is ascending with an upward acceleration \( a \).
2. **Initial velocity of the ball** (with respect to the man): \( v \).
3. **Acceleration of the ball** (with respect to the ground): The effective acceleration acting on the ball when thrown upwards is \( g - a \) (since the lift is accelerating upwards).
4. **Time of flight**: The ball is caught after \( t_1 \) seconds.

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the displacement of the ball,
- \( u = v \) (initial velocity),
- \( a = - (g - a) \) (acceleration acting on the ball),
- \( t = t_1 \).

The displacement of the ball can be expressed as:
\[ s = vt_1 - \frac{1}{2} (g - a) t_1^2 \]

### Step 2: Analyze the second scenario (Lift descending)

1. **Given**: The lift is now descending with the same upward acceleration \( a \) acting downwards.
2. **Initial velocity of the ball** (with respect to the man): \( v \).
3. **Acceleration of the ball** (with respect to the ground): The effective acceleration acting on the ball when thrown upwards is \( g + a \) (since the lift is accelerating downwards).
4. **Time of flight**: The ball is caught after \( t_2 \) seconds.

Using the same equation of motion:
\[ s = vt_2 - \frac{1}{2} (g + a) t_2^2 \]

### Step 3: Set the displacements equal

Since the ball returns to the man, the displacement \( s \) in both scenarios must be equal:
\[ vt_1 - \frac{1}{2} (g - a) t_1^2 = vt_2 - \frac{1}{2} (g + a) t_2^2 \]

### Step 4: Rearranging the equation

Rearranging gives us:
\[ vt_1 - vt_2 = \frac{1}{2} (g - a) t_1^2 - \frac{1}{2} (g + a) t_2^2 \]

This can be simplified to:
\[ v(t_1 - t_2) = \frac{1}{2} [(g - a) t_1^2 - (g + a) t_2^2] \]

### Step 5: Solve for \( v \)

From the above equation, we can express \( v \):
\[ v = \frac{1}{2(t_1 - t_2)} [(g - a) t_1^2 - (g + a) t_2^2] \]

### Step 6: Find the relationship between \( t_1 \) and \( t_2 \)

To find the relationship between \( t_1 \) and \( t_2 \), we can analyze the expressions derived from the equations of motion for both scenarios.

### Final Result

After solving the equations, we find:
1. The acceleration of the lift can be expressed as:
\[ a = \frac{g(t_2 - t_1)}{t_1 + t_2} \]
2. The velocity of the ball relative to the man can be expressed as:
\[ v = \frac{2v}{t_1 + t_2} \]