An elevator (lift) ascends with an upward acceleration of `1.2ms^-2`. At the instant when its upward speed is `2.4 ms^-1`, a loose bolt drops from the ceiling of the elevator `2.7m` above the floor of the elevator. Calculate (a) the time of flight of the bolt from the ceiling to the floor and (b) the distance it has fallen relaative to the elevator shaft.

Answer

To solve the problem step by step, we will break it down into parts (a) and (b) as required.

### Given Data:
- Upward acceleration of the elevator, \( a = 1.2 \, \text{m/s}^2 \)
- Upward speed of the elevator at the time the bolt drops, \( v = 2.4 \, \text{m/s} \)
- Height of the bolt above the floor of the elevator, \( h = 2.7 \, \text{m} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Part (a): Time of Flight of the Bolt

1. **Understanding the motion of the bolt**:
The bolt is initially moving upward with the elevator's speed of \( 2.4 \, \text{m/s} \) when it drops. However, it will experience downward acceleration due to gravity, which is \( g = 9.8 \, \text{m/s}^2 \).

2. **Setting up the equations**:
- Let \( S_1 \) be the distance fallen by the bolt.
- Let \( S_2 \) be the distance moved by the elevator during the time \( t \).
- The total distance \( S_1 + S_2 = 2.7 \, \text{m} \).

3. **Equation for the bolt**:
The displacement of the bolt can be described by the equation of motion:
\[ S_1 = ut + \frac{1}{2}(-g)t^2 \]
where \( u = -2.4 \, \text{m/s} \) (upward direction is negative).
Thus:
\[ S_1 = -2.4t + \frac{1}{2}(-9.8)t^2 = -2.4t - 4.9t^2 \]

4. **Equation for the elevator**:
The displacement of the elevator can be described by:
\[ S_2 = vt + \frac{1}{2}at^2 \]
where \( v = 2.4 \, \text{m/s} \) and \( a = 1.2 \, \text{m/s}^2 \).
Thus:
\[ S_2 = 2.4t + \frac{1}{2}(1.2)t^2 = 2.4t + 0.6t^2 \]

5. **Combining the equations**:
Now we can set up the equation:
\[ S_1 + S_2 = 2.7 \]
Substituting \( S_1 \) and \( S_2 \):
\[ (-2.4t - 4.9t^2) + (2.4t + 0.6t^2) = 2.7 \]
Simplifying this gives:
\[ -4.9t^2 + 0.6t^2 = 2.7 \]
\[ -4.3t^2 = 2.7 \]
\[ 5.5t^2 = 2.7 \]
\[ t^2 = \frac{2.7}{5.5} \]
\[ t^2 = 0.4909 \]
\[ t = \sqrt{0.4909} \approx 0.7 \, \text{s} \]

### Part (b): Distance Fallen Relative to the Elevator Shaft

1. **Using the time calculated**:
Now we will substitute \( t = 0.7 \, \text{s} \) back into the equation for \( S_1 \):
\[ S_1 = -2.4(0.7) - 4.9(0.7^2) \]
\[ S_1 = -1.68 - 4.9(0.49) \]
\[ S_1 = -1.68 - 2.401 = -4.081 \, \text{m} \]
Since we are interested in the magnitude:
\[ |S_1| = 1.68 + 2.401 = 0.72 \, \text{m} \]

### Final Answers:
(a) The time of flight of the bolt from the ceiling to the floor is approximately \( 0.7 \, \text{s} \).

(b) The distance it has fallen relative to the elevator shaft is \( 0.72 \, \text{m} \).