An elevator car whose floor to ceiling distance is equal to `2.7 m` starts ascending with constant acceleration `1.2 m//s^(2)`, 2 sec after the start a bolt begins falling from the ceiling of the car. Answer the following question `(g=9.8 m//s^(2))`
The bolt's free fall time is

Options

A. `sqrt((2.7)/(9.8))s`
B. `sqrt((5.4)/(9.8))s`
C. `sqrt((5.4)/(8.6))s`
D. `sqrt((5.4)/(11))s`

Explanation

To solve the problem, we need to analyze the motion of the bolt falling from the ceiling of the elevator car. The elevator is accelerating upwards, which affects the effective acceleration experienced by the bolt.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Height of the elevator (distance from floor to ceiling) \( s = 2.7 \, \text{m} \)
- Acceleration of the elevator \( a = 1.2 \, \text{m/s}^2 \)
- Time before the bolt falls \( t_0 = 2 \, \text{s} \)
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)

2. **Determine the Effective Acceleration:**
When the bolt starts falling, it experiences two accelerations: the gravitational pull downwards and the upward acceleration of the elevator. The effective acceleration \( a_{\text{net}} \) acting on the bolt is:
\[ a_{\text{net}} = g + a = 9.8 \, \text{m/s}^2 + 1.2 \, \text{m/s}^2 = 11 \, \text{m/s}^2 \]

3. **Use the Second Equation of Motion:**
The second equation of motion states:
\[ s = ut + \frac{1}{2} a_{\text{net}} t^2 \]
Here, the initial velocity \( u = 0 \) (since the bolt starts falling from rest), so the equation simplifies to:
\[ s = \frac{1}{2} a_{\text{net}} t^2 \]

4. **Substitute the Known Values:**
Substitute \( s = 2.7 \, \text{m} \) and \( a_{\text{net}} = 11 \, \text{m/s}^2 \) into the equation:
\[ 2.7 = \frac{1}{2} \cdot 11 \cdot t^2 \]
Simplifying this gives:
\[ 2.7 = 5.5 t^2 \]

5. **Solve for \( t^2 \):**
Rearranging the equation for \( t^2 \):
\[ t^2 = \frac{2.7}{5.5} \]

6. **Calculate \( t \):**
\[ t^2 = \frac{2.7}{5.5} \approx 0.4909 \]
Taking the square root to find \( t \):
\[ t \approx \sqrt{0.4909} \approx 0.7 \, \text{s} \]

### Final Answer:
The bolt's free fall time is approximately \( 0.7 \, \text{s} \).

---