(i) For ball A, using the equation of motion:

\(s_A = \frac{1}{2} g \times 2.5^2 = 31.25 \text{ m}\)

For ball B, using the equation of motion:

\(s_B = 20 \times 1.5 - \frac{1}{2} g \times 1.5^2 = 18.75 \text{ m}\)

Equating the heights:

\(\frac{1}{2} g \times 2.5^2 + 20 \times 1.5 - \frac{1}{2} g \times 1.5^2 = 50 \text{ m}\)

Thus, the height of the tower is 50 m.

(ii) For ball A, using \(s = \frac{1}{2} g t_A^2\):

\(50 = 0.5 g t_A^2 \Rightarrow t_A = 3.16 \text{ s}\)

For ball B, time in motion:

\(t_B = \sqrt{10} - 1 = 2.16 \text{ s}\)

To the top, using \(v^2 = u^2 - 2gs_B\):

\(0^2 = 20^2 - 2gs_B \Rightarrow s_B = 20 \text{ m}\)

To the top, using \(v = u + at\):

\(0 = 20 - gt_B \Rightarrow t_B = 2 \text{ s}\)

Downwards, using \(s = \frac{1}{2} g (0.16)^2\):

\(s_B = 0.13 \text{ m}\)

Total distance for B:

\(20 + 0.13 = 20.1 \text{ m}\)