The height of water in a harbour at time $t$ hours after midnight on a certain day is modelled by the formula $h = 4.8 \sin k\pi t + 5.1$ where $0 \le t \le 24$.

1. Find the maximum level of the tide

The maximum value of $\sin k\pi t$ is $1$.
Maximum level $= 4.8(1) + 5.1 = 9.9$ m

2. Find the minimum level of the tide

The minimum value of $\sin k\pi t$ is $-1$.
Minimum level $= 4.8(-1) + 5.1 = 0.3$ m

1. Determine the period of one tide cycle in hours.

Since there are 2 tide cycles in 24 hours, the period of one cycle is:
$\text{Period} = \frac{24}{2} = 12 \text{ hours}$

2. Relate the period to the coefficient of $t$ in the sine function and solve for $k$.

The period of $y = a \sin(bt) + c$ is $\frac{2\pi}{b}$. Here, $b = k\pi$.
$\frac{2\pi}{k\pi} = 12$
$\frac{2}{k} = 12$
$12k = 2$
$k = \frac{2}{12} = \frac{1}{6} \text{ (shown)}$

1. Set up the inequality for the height of the water

We require $h > 2$.
$4.8 \sin\left(\frac{\pi}{6} t\right) + 5.1 > 2$
$4.8 \sin\left(\frac{\pi}{6} t\right) > -3.1$
$\sin\left(\frac{\pi}{6} t\right) > -\frac{31}{48}$

2. Find the basic angle and the boundary values for the angle

Let $\theta = \frac{\pi}{6} t$. We solve $\sin \theta = -\frac{31}{48}$ for $0 \le \theta \le 2\pi$.
Basic angle $\alpha = \sin^{-1}\left(\frac{31}{48}\right) \approx 0.7017$ rad.
Since $\sin \theta$ is negative, $\theta$ is in the 3rd and 4th quadrants:
$\theta_1 = \pi + 0.7017 = 3.8433$ rad
$\theta_2 = 2\pi - 0.7017 = 5.5815$ rad

3. Solve for $t$ and state the range of times

For $\theta_1$:
$\frac{\pi}{6} t_1 = 3.8433 \implies t_1 = \frac{6}{\pi}(3.8433) \approx 7.340$
For $\theta_2$:
$\frac{\pi}{6} t_2 = 5.5815 \implies t_2 = \frac{6}{\pi}(5.5815) \approx 10.660$

From the sine graph, $\sin\left(\frac{\pi}{6} t\right) > -\frac{31}{48}$ when $0 \le t < 7.340$ and $10.660 < t \le 12$.
Thus, the range of times is $0 \le t < 7.34$ and $10.66 < t \le 12$.