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 situation of a coin being dropped in two different scenarios: when the elevator is stationary and when it is moving uniformly.

### Step-by-Step Solution:

1. **Understanding the Scenarios**:
- **Scenario 1**: The elevator is stationary. When the person drops the coin, it falls under the influence of gravity. The time taken for the coin to reach the floor of the elevator is denoted as \( t_1 \).
- **Scenario 2**: The elevator is moving uniformly (either upwards or downwards). The time taken for the coin to reach the floor of the elevator in this case is denoted as \( t_2 \).

2. **Applying the Equations of Motion**:
- For the stationary elevator, we can use the second equation of motion:
\[ h = \frac{1}{2} g t_1^2 \]
Rearranging gives:
\[ t_1 = \sqrt{\frac{2h}{g}} \]

- For the moving elevator, we consider that the elevator is moving with a constant velocity \( v \). In this case, we can analyze it from the perspective of an observer inside the elevator. The coin is still falling under gravity, and the time taken to reach the floor is:
\[ h = \frac{1}{2} g t_2^2 \]
Rearranging gives:
\[ t_2 = \sqrt{\frac{2h}{g}} \]

3. **Comparing the Times**:
- Since both equations for \( t_1 \) and \( t_2 \) yield the same expression:
\[ t_1 = \sqrt{\frac{2h}{g}} \quad \text{and} \quad t_2 = \sqrt{\frac{2h}{g}} \]
- This shows that \( t_1 = t_2 \).

4. **Conclusion**:
- Therefore, the time taken for the coin to reach the floor of the elevator is the same in both cases, regardless of whether the elevator is stationary or moving uniformly. The correct answer is that \( t_1 = t_2 \).

### Final Answer:
- \( t_1 = t_2 \)