Let the upward direction be positive. For the first particle projected upwards:

Using the equation of motion: \(s_1 = ut - \frac{1}{2}gt^2\)

\(s_1 = 12.5t - \frac{1}{2}gt^2\)

For the second particle released from rest:

Using the equation of motion: \(s_2 = \frac{1}{2}gt^2\)

Since the total distance between the particles is 10 m, we have:

\(s_1 + s_2 = 10\)

Substituting the expressions for \(s_1\) and \(s_2\):

\(12.5t - \frac{1}{2}gt^2 + \frac{1}{2}gt^2 = 10\)

\(12.5t = 10\)

\(t = 0.8 \text{ s}\)

Substitute \(t = 0.8\) back into the equation for \(s_1\):

\(s_1 = 12.5 \times 0.8 - \frac{1}{2} \times 9.8 \times (0.8)^2\)

\(s_1 = 10 - 3.2\)

\(s_1 = 6.8 \text{ m}\)

Therefore, the height above O at which the particles meet is 6.8 m.

(i) For \(P_1\), using \(s = ut - \frac{1}{2}gt^2\), we have:

\(25 = 30t - 5t^2\)

Solving \(5t^2 - 30t + 25 = 0\) gives \((t-1)(t-5) = 0\), so \(t = 1\) or \(t = 5\).

Thus, \(P_1\) is higher than the tower for \(1 < t < 5\).

(ii) For \(P_1\) and \(P_2\), using \(s_1 = 30t - 5t^2\) and \(s_2 = 10t - 5t^2\), and \(s_1 = s_2 + 25\):

\(30t - 10t = 25\)

\(t = 1.25\)

Velocities: \(v_1 = 30 - 10 \times 1.25 = 17.5\) m s\(^{-1}\), \(v_2 = 10 - 10 \times 1.25 = -2.5\) m s\(^{-1}\).

(iii) Using \(t_{up} = 3\) and \(t_{equal} = 1.25\), the time is \(1.75\) s or \(1.25 < t < 3\).

1. To find the speed after 2 s, use the formula for velocity under constant acceleration: \(v = u + gt\). Since the stone is released from rest, \(u = 0\). Thus, \(v = 0 + 10 \times 2 = 20 \text{ m s}^{-1}\).
2. To find the time taken to fall 45 m, use the formula \(s = \frac{1}{2}gt^2\). Solving \(45 = \frac{1}{2} \times 10 \times t^2\) gives \(t^2 = 9\), so \(t = 3 \text{ s}\).
3. To find the distance fallen from 30 m s^-1 to 40 m s^-1, use \(v^2 = u^2 + 2gs\). Substituting \(40^2 = 30^2 + 2 \times 10 \times s\), we get \(1600 = 900 + 20s\). Solving gives \(s = 35 \text{ m}\).

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

\(v = u - gt\)

where \(v = 0\) at maximum height, \(u = 5 \text{ m/s}\), and \(g = 9.8 \text{ m/s}^2\). Solving \(0 = 5 - 9.8t\) gives \(t = \frac{5}{9.8}\).

The height \(h\) is given by:

\(h = ut - \frac{1}{2}gt^2\)

Substitute \(t = \frac{5}{9.8}\) to find \(h_A\).

For B, use the same formula with \(u = 8 \text{ m/s}\) and the same \(t\). Calculate \(h_B\).

The difference in heights is \(h_B - h_A = 1.5 \text{ m}\).

(ii) To find when B is 0.9 m above A, solve:

\(h_B - h_A = 0.9\)

Using \(h = ut - \frac{1}{2}gt^2\) for both particles, set up the equation:

\(8t - \frac{1}{2}gt^2 - (5t - \frac{1}{2}gt^2) = 0.9\)

Simplify to \(3t = 0.9\), giving \(t = 0.3 \text{ s}\).

Substitute \(t = 0.3\) into \(h_A = 5t - \frac{1}{2}gt^2\) to find \(h_A = 1.05 \text{ m}\).

(a) To find the initial speed, use the equation of motion:

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

At the maximum height, \(v = 0\), \(a = -g\), and \(s = 45\) m. Thus,

\(0 = u^2 - 2g \times 45\)

Solving for \(u\), we get:

\(u^2 = 2g \times 45\)

\(u = \sqrt{2 \times 9.8 \times 45} = 30 \text{ m s}^{-1}\)

(b) To find the total time for which the speed is at least 10 m s^-1, use the equation:

\(v = u + at\)

Set \(v = 10 \text{ m s}^{-1}\), \(u = 30 \text{ m s}^{-1}\), and \(a = -g\):

\(10 = 30 - gt\)

\(gt = 20\)

\(t = \frac{20}{9.8} \approx 2 \text{ s}\)

The particle takes 2 s to reach 10 m s^-1 on the way up and another 2 s on the way down, so the total time is:

\(2 \times 2 = 4 \text{ s}\)