(i) For particle A, the height at time t is given by:

\(h_A = 20t - \frac{1}{2} \times 10t^2\)

For particle B, the height at time t is:

\(h_B = 40 - \frac{1}{2} \times 10(t-1)^2\)

Setting \(h_A = h_B\) to find the collision point:

\(20t - \frac{1}{2} \times 10t^2 = 40 - \frac{1}{2} \times 10(t-1)^2\)

Simplifying gives:

\(10t - 35 = 0\)

Solving for \(t\):

\(t = 3.5\)

Substitute \(t = 3.5\) into \(h_A\) or \(h_B\):

\(h = 20 \times 3.5 - \frac{1}{2} \times 10 \times (3.5)^2 = 8.75 \text{ m}\)

(ii) The speed of particle A at collision:

\(v_A = 20 - gt = 20 - 10 \times 3.5 = -15 \text{ m s}^{-1}\)

The speed of particle B at collision:

\(v_B = -g(t-1) = -10 \times 2.5 = -25 \text{ m s}^{-1}\)

Difference in speeds:

\(|v_A - v_B| = |-15 - (-25)| = 10 \text{ m s}^{-1}\)