(i) The time for particle P to reach the ground is given by solving the equation for vertical motion:

\(17t - 5t^2 = 0\)

Solving gives \(t = 0\) or \(t = 3.4\) seconds. The time for particle Q to reach the ground is given by:

\(7(t - T) - 5(t - T)^2 = 0\)

Solving gives \(t - T = 0\) or \(t - T = 1.4\) seconds. Since both particles reach the ground at the same time, \(t = 3.4\) and \(t - T = 1.4\), so \(T = 2\) seconds.

(ii) At the instant when P is 5 m higher than Q, we have:

\(17(t + 2) - 5(t + 2)^2 - (7t - 5t^2) = 5\)

Solving gives \(t = 0.9\) or \(t = 2.9\) seconds. Using \(t = 0.9\), the velocities are:

For P: \(v_P = 17 - 10(t + 2) = 12 \text{ m s}^{-1}\)

For Q: \(v_Q = 7 - 10t = 2 \text{ m s}^{-1}\)

Both velocities are directed vertically downwards.