A particle is thrown up inside a stationary lift of sufficient height. The time of flight is T. Now it is thrown again with same initial speed `v_0` with respect to lift. At the time of second throw, lift is moving up with speed `v_0` and uniform acceleration g upward (the acceleration due to gravity). The new time of flight is:

Options

A. `T/4`
B. `T/2`
C. `T`
D. `2T`

Answer

To solve the problem, we need to analyze the motion of the particle in both scenarios: when the lift is stationary and when the lift is moving upward with a certain velocity and acceleration.

### Step-by-Step Solution:

**Step 1: Analyze the first case (Lift is stationary)**

In the first scenario, the particle is thrown upwards with an initial speed \( v_0 \) from a stationary lift. The time of flight \( T \) can be derived using the following kinematic equation:

\[ s = ut + \frac{1}{2} a t^2 \]

Where:
- \( s \) = displacement (which is 0, as the particle returns to the same height)
- \( u = v_0 \) (initial velocity)
- \( a = -g \) (acceleration due to gravity, acting downwards)
- \( t = T \) (time of flight)

Substituting these values into the equation gives:

\[ 0 = v_0 T - \frac{1}{2} g T^2 \]

Rearranging this, we find:

\[ v_0 T = \frac{1}{2} g T^2 \]

Dividing both sides by \( T \) (assuming \( T \neq 0 \)):

\[ v_0 = \frac{1}{2} g T \]

From this, we can express \( T \):

\[ T = \frac{2v_0}{g} \]

**Step 2: Analyze the second case (Lift is moving upward)**

In the second scenario, the lift is moving upward with a speed \( v_0 \) and has an upward acceleration of \( g \). We need to find the new time of flight \( T' \).

In this case, the effective acceleration acting on the particle (relative to the lift) is:

\[ a' = -g - g = -2g \]

The kinematic equation remains the same:

\[ s = ut + \frac{1}{2} a' t^2 \]

Where:
- \( s = 0 \) (the particle returns to the same height)
- \( u = v_0 \) (initial velocity)
- \( a' = -2g \) (effective acceleration)

Substituting these values gives:

\[ 0 = v_0 T' + \frac{1}{2} (-2g) (T')^2 \]

This simplifies to:

\[ 0 = v_0 T' - g (T')^2 \]

Rearranging gives:

\[ g (T')^2 = v_0 T' \]

Dividing both sides by \( T' \) (assuming \( T' \neq 0 \)):

\[ g T' = v_0 \]

Thus, we can express \( T' \):

\[ T' = \frac{v_0}{g} \]

**Step 3: Relate \( T' \) to \( T \)**

From our earlier result for \( T \):

\[ T = \frac{2v_0}{g} \]

Now, we can express \( T' \) in terms of \( T \):

\[ T' = \frac{v_0}{g} = \frac{1}{2} \left( \frac{2v_0}{g} \right) = \frac{1}{2} T \]

### Final Answer:

The new time of flight \( T' \) is:

\[ T' = \frac{T}{2} \]