(i) To find the time taken for P to reach its greatest height, use the equation of motion:

\(v = u + at\)

where \(v = 0\) (at the greatest height), \(u = 25\) m/s, and \(a = -g = -10\) m/s². Solving for \(t\):

\(0 = 25 - 10t\)

\(t = 2.5 \text{ s}\)

(ii) To find the length of time for which P is higher than 23 m above the ground, use the equation:

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

where \(s = 20\) m (since 23 m above the ground is 20 m above the projection point), \(u = 25\) m/s, and \(a = -10\) m/s². Solving the quadratic equation:

\(20 = 25t - 5t^2\)

Rearranging gives:

\(5t^2 - 25t + 20 = 0\)

Solving this quadratic equation gives \(t = 1\) s and \(t = 4\) s. The time interval is:

\(4 - 1 = 3 \text{ s}\)

(iii) To find h such that P is higher than h m above the ground for 1 second, note that the maximum height is reached at 2.5 s. Therefore, P reaches h after 2 s. Using:

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

where \(u = 25\) m/s, \(a = -10\) m/s², and \(t = 2\) s:

\(h - 3 = 25 \times 2 - 5 \times 2^2\)

\(h - 3 = 50 - 20\)

\(h = 33 \text{ m}\)