(a) To find the initial speed u, we use the equation for velocity in uniformly accelerated motion:

\(v = u + at\)

At the greatest height, the final velocity \(v = 0\), the acceleration \(a = -10 \text{ m/s}^2\) (due to gravity), and the time \(t = 3 \text{ s}\). Substituting these values, we get:

\(0 = u - 10 \times 3\)

\(u = 30 \text{ m/s}\)

(b) To find the greatest height, we use the equation:

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

Substituting \(v = 0\), \(u = 30 \text{ m/s}\), \(a = -10 \text{ m/s}^2\), we get:

\(0 = 30^2 - 2 \times 10 \times s\)

\(0 = 900 - 20s\)

\(20s = 900\)

\(s = 45 \text{ m}\)

Therefore, the greatest height is 45 m.