The depth of water, $h$ metres, in a shallow port on a particular day is modelled by the formula $h = 7 + 2\sin\left(\frac{\pi}{6}t\right),$ where $t$ is the number of hours after midnight.

1. Calculate the period using the coefficient of $t$

$\text{Period} = \frac{2\pi}{\frac{\pi}{6}} = 12$

1. Identify the condition for the minimum depth

The depth $h = 7 + 2\sin\left(\frac{\pi}{6}t\right)$ is at its lowest when $\sin\left(\frac{\pi}{6}t\right) = -1$.

2. Solve for $t$

$\frac{\pi}{6}t = \frac{3\pi}{2}$
$t = \frac{3\pi}{2} \times \frac{6}{\pi} = 9$

3. Convert $t$ to a time of day

$t = 9$ corresponds to 9 hours after midnight, which is 0900 or 9.00 am.

1. Determine the time when unloading finishes.

The boat docks at 5.30 am and takes 3 hours to unload.
Unloading finishes at 8.30 am, which corresponds to $t = 8.5$.

2. Set up the equation for the required depth.

We need $h \geqslant 5.6$.
$7 + 2\sin\left(\frac{\pi}{6}t\right) = 5.6$
$2\sin\left(\frac{\pi}{6}t\right) = -1.4$
$\sin\left(\frac{\pi}{6}t\right) = -0.7$

3. Solve for the basic angle and find the relevant value of $t$.

Basic angle $\alpha = \sin^{-1}(0.7) \approx 0.7754$ rad.
Since $\sin\left(\frac{\pi}{6}t\right)$ is negative, $\frac{\pi}{6}t$ is in the 3rd or 4th quadrant.
For the 3rd quadrant: $\frac{\pi}{6}t = \pi + 0.7754 \implies t \approx 7.48$ (which is before 8.5, so rejected).
For the 4th quadrant: $\frac{\pi}{6}t = 2\pi - 0.7754 \approx 5.5078$
$t = \frac{6}{\pi} \times 5.5078 \approx 10.519$

4. Convert the time to hours and minutes.

$t = 10.519$ hours corresponds to 10 hours and $0.519 \times 60 \approx 31.14$ minutes.
To the nearest minute, the time is 10:31 am.