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 analyze the motion of the coin released from the lift. Here are the steps to find the time after which the coin will strike the lift:

### Step 1: Understand the motion of the lift and the coin
The lift is moving downwards with an acceleration of \( a = 9 \, \text{m/s}^2 \). The coin is released from a height of \( h = 2 \, \text{m} \) above the floor of the lift. The gravitational acceleration is given as \( g = 10 \, \text{m/s}^2 \).

### Step 2: Determine the effective acceleration of the coin
When the coin is released, it experiences two accelerations:
1. Downward gravitational acceleration \( g = 10 \, \text{m/s}^2 \).
2. Upward pseudo force due to the lift's downward acceleration \( a = 9 \, \text{m/s}^2 \).

The effective acceleration of the coin relative to the lift can be calculated as:
\[ a_{\text{effective}} = g - a = 10 \, \text{m/s}^2 - 9 \, \text{m/s}^2 = 1 \, \text{m/s}^2 \]
This means the coin accelerates downwards at \( 1 \, \text{m/s}^2 \) relative to the lift.

### Step 3: Apply the kinematic equation
We will use the kinematic equation for uniformly accelerated motion:
\[ S = ut + \frac{1}{2} a t^2 \]
Where:
- \( S = 2 \, \text{m} \) (the distance the coin falls),
- \( u = 0 \, \text{m/s} \) (initial velocity of the coin),
- \( a = 1 \, \text{m/s}^2 \) (effective acceleration),
- \( t \) is the time in seconds.

Substituting the known values into the equation:
\[ 2 = 0 \cdot t + \frac{1}{2} \cdot 1 \cdot t^2 \]
This simplifies to:
\[ 2 = \frac{1}{2} t^2 \]

### Step 4: Solve for time \( t \)
Multiply both sides by 2 to eliminate the fraction:
\[ 4 = t^2 \]
Taking the square root of both sides gives:
\[ t = 2 \, \text{s} \]

### Conclusion
The time after which the coin will strike the lift is \( t = 2 \, \text{seconds} \).