(i) To find when the particles are travelling in opposite directions, we need to determine when one particle is moving upwards and the other downwards. The time to reach maximum height for P is given by setting the velocity to zero:

\(0 = 12 - gt \Rightarrow t = \frac{12}{g} \approx 1.2 \text{ s}\)

For Q, \(0 = 7 - gt \Rightarrow t = \frac{7}{g} \approx 0.7 \text{ s}\)

Thus, the particles are travelling in opposite directions for \(0.7 < t < 1.2\).

(ii) Given \(3h_P = 8h_Q\), we use the equations for height:

\(h_P = 12t - \frac{1}{2}gt^2\)

\(h_Q = 7t - \frac{1}{2}gt^2\)

Substitute into the given condition:

\(3(12t - \frac{1}{2}gt^2) = 8(7t - \frac{1}{2}gt^2)\)

Simplifying gives:

\(36t - 1.5gt^2 = 56t - 4gt^2\)

\(2.5gt^2 = 20t\)

\(t = \frac{8}{g}\)

Using \(v = u - gt\) for velocities:

\(v_P = 12 - gt\)

\(v_Q = 7 - gt\)

Substitute \(t = \frac{8}{g}\):

\(v_P = 12 - 8 = 4 \text{ m s}^{-1}\)

\(v_Q = 7 - 8 = -1 \text{ m s}^{-1}\)

(i) To find the time for which P's height is greater than 15 m, use the equation for vertical motion:

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

where \(u = 20\) m s^-1 and \(g = 10\) m s^-2. Set \(h = 15\) m:

\(15 = 20t - 5t^2\)

Solve the quadratic equation:

\(5t^2 - 20t + 15 = 0\)

Divide by 5:

\(t^2 - 4t + 3 = 0\)

Factorize:

\((t - 1)(t - 3) = 0\)

So, \(t = 1\) or \(t = 3\). The duration is \(3 - 1 = 2\) seconds.

(ii) To find the velocities when P and Q are at the same height, equate their height equations:

\(20t - 5t^2 = 25(t - 0.4) - 5(t - 0.4)^2\)

Expand and simplify:

\(20t - 5t^2 = 25t - 10 - 5(t^2 - 0.8t + 0.16)\)

\(20t - 5t^2 = 25t - 10 - 5t^2 + 4t - 0.8\)

Combine terms:

\(20t = 29t - 10.8\)

\(9t = 10.8\)

\(t = 1.2\) (or \(t = 0.8\))

Use \(v = u - gt\) to find velocities:

For P:

\(v_P = 20 - 10 \times 1.2 = 8 \text{ m s}^{-1}\)

For Q:

\(v_Q = 25 - 10 \times (1.2 - 0.4) = 17 \text{ m s}^{-1}\)

(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.

(i) For particle P, using the equation of motion under gravity:

\(0 = V - gt\)

\(At t = 2 s, the velocity is zero, so:\)

\(0 = V - g \times 2\)

Solving for V gives:

\(V = 2g\)

\(Assuming g = 10 m s^-2, we find:\)

\(V = 20 \text{ m s}^{-1}\)

(ii) For particle Q, using the equation of motion:

\(v = u + gt\)

\(Since Q starts from rest, u = 0, and at t = 4 s:\)

\(v = 0 + 10 \times 4 = 40 \text{ m s}^{-1}\)

(iii) The height h from which Q was dropped can be found using:

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

\(Substituting t = 4 s:\)

\(h = \frac{1}{2} \times 10 \times 4^2 = 80 \text{ m}\)

(i) Using the equation of motion \(v^2 = u^2 + 2as\) where initial velocity \(u = 0\), acceleration \(a = 10 \text{ m/s}^2\), and distance \(s = 1.25 \text{ m}\):

\(v^2 = 2 \times 10 \times 1.25\)

\(v^2 = 25\)

\(v = 5 \text{ m/s}\)

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

\(1.25 = \frac{1}{2} \times 10 \times t^2\)

\(t^2 = 0.25\)

\(t = 0.5 \text{ s}\)

(ii) The velocity after passing through A is given by integrating the acceleration:

\(v = \int (10 - 0.3t) \, dt = 10t - 0.15t^2 + C\)

At \(t = 0\), \(v = 5 \text{ m/s}\), so \(C = 5\).

\(v = 10t - 0.15t^2 + 5\)

Integrating to find the distance:

\(x = \int (10t - 0.15t^2 + 5) \, dt = 5t^2 - 0.05t^3 + 5t\)

Substitute \(t = 2.5\) (since \(t = 3 - 0.5\)):

\(x = 5 \times 2.5^2 - 0.05 \times 2.5^3 + 5 \times 2.5\)

\(x = 42.97 \text{ m}\)

Total distance \(OP = 1.25 + 42.97 = 44.2 \text{ m}\)