(i) The maximum height occurs when \(\cos kt = -1\). Substituting this into the formula gives:

\(h = 60(1 - (-1)) = 60 \times 2 = 120 \text{ m}\)

(ii) Given that one complete revolution takes 30 minutes, \(kt = 2\pi\) when \(t = 30\). Therefore,

\(30k = 2\pi\)

\(k = \frac{2\pi}{30} = \frac{\pi}{15}\)

(iii) To find the time when the height is above 90 m, set \(h = 90\):

\(90 = 60(1 - \cos kt)\)

\(1 - \cos kt = \frac{90}{60} = 1.5\)

\(\cos kt = -0.5\)

The solutions for \(\cos kt = -0.5\) are \(kt = \frac{2\pi}{3}\) or \(kt = \frac{4\pi}{3}\).

Solving for \(t\):

\(t = \frac{2\pi}{3k} = \frac{2\pi}{3 \times \frac{\pi}{15}} = 10 \text{ minutes}\)

\(t = \frac{4\pi}{3k} = \frac{4\pi}{3 \times \frac{\pi}{15}} = 20 \text{ minutes}\)

Thus, the passenger is above 90 m for 10 minutes.