A person standing oin the floor of an elevator drops as coin. The coin reaches the floor of the elevator in a time `t_1` if the elevator is stationary and in the `t_2` if it is moving uniformly. Then

Options

A. `t_1=t_2`
B. `t_1ltt_2`
C. `t_1gtt_2`
D. `t_1ltt_2 or t_1gtt_2` depending on whether the lift is going up or down.

Answer

To solve the problem, we need to analyze the two scenarios: when the elevator is stationary and when it is moving uniformly. We will derive the relationship between the times \( t_1 \) and \( t_2 \) taken by the coin to reach the floor of the elevator in both cases.

### Step-by-Step Solution:

1. **Understanding the Situation**:
- When the elevator is stationary, the coin is dropped from a height \( h \) and falls freely under gravity.
- When the elevator is moving uniformly, the coin is still dropped from the same height \( h \), but the elevator is moving at a constant velocity.

2. **Case 1: Elevator is Stationary**:
- The time \( t_1 \) taken for the coin to fall to the floor can be derived from the equation of motion under gravity:
\[ h = \frac{1}{2} g t_1^2 \]
- Rearranging this gives:
\[ t_1^2 = \frac{2h}{g} \quad \Rightarrow \quad t_1 = \sqrt{\frac{2h}{g}} \]

3. **Case 2: Elevator is Moving Uniformly**:
- In this case, the elevator is moving with a constant velocity \( V \). When the coin is dropped, it has the same initial velocity \( V \) as the elevator.
- The relative motion can be analyzed from the frame of the elevator. For an observer in the elevator, the coin appears to fall from the same height \( h \) as before.
- The time \( t_2 \) taken for the coin to reach the floor of the elevator is also given by:
\[ h = \frac{1}{2} g t_2^2 \]
- Rearranging this gives:
\[ t_2^2 = \frac{2h}{g} \quad \Rightarrow \quad t_2 = \sqrt{\frac{2h}{g}} \]

4. **Comparing the Two Times**:
- From both cases, we have:
\[ t_1 = \sqrt{\frac{2h}{g}} \quad \text{and} \quad t_2 = \sqrt{\frac{2h}{g}} \]
- Therefore, we conclude that:
\[ t_1 = t_2 \]

### Conclusion:
The time taken for the coin to reach the floor of the elevator is the same in both cases, thus:
\[ t_1 = t_2 \]