A coin is released inside a lift at a height of 2m from the floor of the lift. The height of the lift is 10 m. The lift is moving with an acceleration of `9 m//s^(2)`downwards. The time after which the coin will strike with the lift is : `(g=10 m//s^(2))`

Options

A. 4s
B. 2s
C. `4/(sqrt(21))s`
D. `2/(sqrt(11)) s`

Explanation

To solve the problem, we need to determine the time it takes for the coin to strike the lift after being released. Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The coin is released from a height of 2 meters above the floor of the lift.
- The lift itself is moving downwards with an acceleration of \(9 \, \text{m/s}^2\).
- The acceleration due to gravity is \(g = 10 \, \text{m/s}^2\).

2. **Determine the Relative Motion:**
- The initial velocity of the coin relative to the lift when it is released is \(u = 0 \, \text{m/s}\) (since it is released and not thrown).
- The effective acceleration of the coin relative to the lift can be calculated as:
\[ a_{\text{relative}} = g - a_{\text{lift}} = 10 \, \text{m/s}^2 - 9 \, \text{m/s}^2 = 1 \, \text{m/s}^2 \]

3. **Use the Equation of Motion:**
- We will use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
- Here, \(s\) is the distance the coin will fall relative to the lift, which is \(2 \, \text{m}\), \(u = 0\), and \(a = 1 \, \text{m/s}^2\).
- Substituting the values into the equation:
\[ 2 = 0 \cdot t + \frac{1}{2} \cdot 1 \cdot t^2 \]
- This simplifies to:
\[ 2 = \frac{1}{2} t^2 \]

4. **Solve for Time \(t\):**
- Rearranging gives:
\[ t^2 = 2 \cdot 2 = 4 \]
- Taking the square root:
\[ t = \sqrt{4} = 2 \, \text{seconds} \]

5. **Conclusion:**
- The time after which the coin will strike the lift is \(2 \, \text{seconds}\).

### Final Answer:
The time after which the coin will strike the lift is **2 seconds**.