An elevator, in which a man is standing, is moving upward with a constant acceleration of `2 m//s^2`. At some instant when speed of elevator is `10 m//s`, the man drops a coin from a height of `1.5 m`. Find the time taken by the coin to reach the floor.

Options

A. `(1)/(sqrt(3)) sec`
B. `(1)/(2) sec`
C. `(1)/(sqrt(2)) sec`
D. `1 sec`

Explanation

To solve the problem of the coin dropped from the elevator, we can follow these steps:

### Step 1: Understand the scenario
The elevator is moving upward with a constant acceleration of \(2 \, \text{m/s}^2\) and has a speed of \(10 \, \text{m/s}\) at the moment the coin is dropped from a height of \(1.5 \, \text{m}\).

### Step 2: Define the motion of the coin
When the coin is dropped, it has an initial velocity equal to the speed of the elevator, which is \(10 \, \text{m/s}\) upward. However, once dropped, the only force acting on the coin is gravity, which accelerates it downward at \(g = 9.8 \, \text{m/s}^2\).

### Step 3: Establish the reference frame
We will consider the upward direction as positive. Therefore:
- The initial velocity of the coin with respect to the ground is \(u = 10 \, \text{m/s}\) (upward).
- The acceleration of the coin due to gravity is \(a = -9.8 \, \text{m/s}^2\) (downward).

### Step 4: Calculate the relative motion
Since the elevator is also moving upward with an acceleration of \(2 \, \text{m/s}^2\), we need to find the relative acceleration of the coin with respect to the elevator.

The acceleration of the elevator is:
- \(a_{\text{elevator}} = 2 \, \text{m/s}^2\) (upward).

Thus, the relative acceleration of the coin with respect to the elevator is:
\[ a_{\text{relative}} = a_{\text{coin}} - a_{\text{elevator}} = -9.8 \, \text{m/s}^2 - 2 \, \text{m/s}^2 = -11.8 \, \text{m/s}^2 \]

### Step 5: Set up the equation of motion
Using the equation of motion for the coin with respect to the elevator:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \(s = -1.5 \, \text{m}\) (the coin moves downward relative to the elevator),
- \(u = 0 \, \text{m/s}\) (the relative velocity of the coin with respect to the elevator is \(0\) when it is dropped),
- \(a = -11.8 \, \text{m/s}^2\).

Substituting the values into the equation:
\[ -1.5 = 0 \cdot t + \frac{1}{2} \cdot (-11.8) \cdot t^2 \]
This simplifies to:
\[ -1.5 = -5.9 t^2 \]
Dividing both sides by \(-5.9\):
\[ t^2 = \frac{1.5}{5.9} \]
Calculating \(t^2\):
\[ t^2 \approx 0.2542 \]
Taking the square root:
\[ t \approx \sqrt{0.2542} \approx 0.5042 \, \text{s} \]

### Step 6: Final answer
Thus, the time taken by the coin to reach the floor of the elevator is approximately \(0.5 \, \text{s}\).

---