(a) Using the equation of motion:

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

where \(u = 25 \text{ m s}^{-1}, a = -g = -9.8 \text{ m s}^{-2}, s = 20 \text{ m}\).

\(v^2 = 25^2 + 2(-9.8)(20)\)

\(v^2 = 625 - 392\)

\(v^2 = 233\)

\(v = \sqrt{233} \approx 15 \text{ m s}^{-1}\)

(b) Using conservation of momentum:

Taking upwards as positive:

\(0.5v_A + 0.3(-32.5) = 0.5(-15) + 0\)

\(0.5v_A - 9.75 = -7.5\)

\(0.5v_A = 2.25\)

\(v_A = 4.5 \text{ m s}^{-1}\) downwards

(c) For particle A:

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

\(20 = 4.5t_A + \frac{1}{2}(9.8)t_A^2\)

Solving gives \(t_A = 1.6 \text{ s}\)

For particle B:

\(20 = \frac{1}{2}(9.8)t_B^2\)

Solving gives \(t_B = 2 \text{ s}\)

Time interval = \(t_B - t_A = 2 - 1.6 = 0.4 \text{ s}\)