(a) To find the height h of the building, we use the equation of motion for the first particle:

Using the equation for velocity, \(v = u + at\), with \(u = 40\), \(v = 0\), and \(a = -g\), we find the time to reach the highest point:

\(40 - gt = 0\) gives \(t = 4\) seconds.

The time to reach the top of the building is \(4 - \frac{1}{2} \times 10 \times 2^2 = 2\) seconds.

Using the equation for displacement, \(s = ut + \frac{1}{2} at^2\), with \(u = 40\), \(a = -g\), and \(t = 2\), we find:

\(h = 40 \times 2 - \frac{1}{2} \times 10 \times 2^2 = 60\) m.

(b) For the first particle, the height above ground is given by:

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

For the second particle, the height above the top of the building is:

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

Setting the heights equal gives:

\(60 + 20(t - 1) - \frac{1}{2} \times 10 \times (t - 1)^2 = 40t - \frac{1}{2} \times 10t^2\).

Solving this equation, we find \(t = 3.5\) seconds.