(i) To find the maximum height, use the equation of motion:

\(v^2 = u^2 + 2as\)

where \(v = 0\) (at maximum height), \(u = 30\) m/s, and \(a = -g\) (acceleration due to gravity, \(g = 9.8\) m/s²).

\(0 = 30^2 + 2(-9.8)s\)

\(s = \frac{900}{19.6} = 45\) m

(ii) To find the time to reach 33.75 m, use:

\(s = ut + \frac{1}{2}at^2\)

\(33.75 = 30t - \frac{1}{2}gt^2\)

\(33.75 = 30t - 4.9t^2\)

\(4.9t^2 - 30t + 33.75 = 0\)

Solving this quadratic equation gives \(t = 1.5\) s (rejecting \(t = 4.5\) s as it is the second time reaching this height).

To find the speed at this time, use:

\(v = u + at\)

\(v = 30 - 9.8 \times 1.5\)

\(v = 15\) m/s