(a) To find the initial speed u, use the equation for maximum height:

\(0 = u^2 - 2 \times 10 \times 20\)

Solving for \(u\), we get:

\(u^2 = 400\)

\(u = 20 \text{ m/s}\)

(b) To find the total time for which P is at least 15 m above the ground, use the equation:

\(15 = 20t - \frac{1}{2} \times 10 \times t^2\)

Rearranging gives:

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

Solving this quadratic equation, we find:

\(t = 1 \text{ or } t = 3\)

The total time is \(3 - 1 = 2 \text{ s}\).

(a) Using the equation of motion: \(s = ut + \frac{1}{2} a t^2\), where \(s = 24 \text{ m}\), \(t = 2 \text{ s}\), and \(a = -g = -9.8 \text{ m s}^{-2}\).

\(24 = u \times 2 - \frac{1}{2} \times 9.8 \times 2^2\)

\(24 = 2u - 19.6\)

\(2u = 43.6\)

\(u = 22 \text{ m s}^{-1}\)

(b) At maximum height, \(v = 0\), using \(v^2 = u^2 + 2as\):

\(0 = 22^2 - 2 \times 9.8 \times s\)

\(s = 24.2 \text{ m}\)

Height down in 1.8 s: \(s = \frac{1}{2} \times 9.8 \times 1.8^2 = 16.2 \text{ m}\)

\(h = 24.2 - 16.2 = 8 \text{ m}\)

Alternative method: Using \(v = u - gt\) with \(u = 22\) and \(v = 0\):

\(0 = 22 - 9.8t\)

\(t = 2.2 \text{ s}\)

\(h = 22 \times (2.2 - 1.8) - \frac{1}{2} \times 9.8 \times (2.2 - 1.8)^2\)

\(h = 8 \text{ m}\)

(a) To find the initial velocity v, we use the equation of motion:

\(v = u + at\)

where \(v = 0\) (velocity at the greatest height), \(a = -g = -10 \text{ m/s}^2\) (acceleration due to gravity), and \(t = 3 \text{ s}\).

Substituting the values, we get:

\(0 = v - 10 \times 3\)

\(v = 30 \text{ m/s}\)

(b) To find the greatest height, we use the equation:

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

where \(v = 0\), \(u = 30 \text{ m/s}\), and \(a = -10 \text{ m/s}^2\).

Substituting the values, we get:

\(0 = 30^2 + 2(-10)s\)

\(0 = 900 - 20s\)

\(20s = 900\)

\(s = 45 \text{ m}\)

Thus, the greatest height is 45 m.

(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.

(a) To find the greatest height above the ground reached by P, we use the equation for vertical motion:

\(v^2 = u^2 - 2gs\)

where \(v = 0\) m/s (at the highest point), \(u = 5\) m/s, and \(g = 9.8\) m/s².

\(0 = 5^2 - 2 \times 9.8 \times s\)

\(s = \frac{25}{19.6} = 1.25\) m

The height above the ground is \(2.8 + 1.25 = 4.05\) m.

(b) To find the length of time for which P is at a height of more than 3.6 m above the ground, we use the equation:

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

Let \(s = 3.6 - 2.8 = 0.8\) m, \(u = 5\) m/s, \(a = -9.8\) m/s².

\(0.8 = 5t - 4.9t^2\)

Solving the quadratic equation \(4.9t^2 - 5t + 0.8 = 0\), we find \(t = 0.2\) s and \(t = 0.8\) s.

The length of time for which P is above 3.6 m is \(0.8 - 0.2 = 0.6\) s.