Using the SUVAT equation for vertical motion: \(s = ut - \frac{1}{2}gt^2\), where \(g = 10 \text{ m s}^{-2}\).

At \(t = 3\) seconds, \(s = 3u - 5 \times 9\).

At \(t = 4\) seconds, \(s = 4u - 5 \times 16\).

Equating the two expressions for \(s\):

\(3u - 45 = 4u - 80\)

Solving for \(u\):

\(3u - 4u = -80 + 45\)

\(-u = -35\)

\(u = 35\)

Substitute \(u = 35\) back into one of the equations for \(s\):

\(s = 3(35) - 45 = 105 - 45 = 60\)

Thus, \(u = 35\) and \(s = 60\).

(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:

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

(i) To find the time taken for P to reach its greatest height, use the equation of motion:

\(v = u + at\)

where \(v = 0\) (at the greatest height), \(u = 25\) m/s, and \(a = -g = -10\) m/s². Solving for \(t\):

\(0 = 25 - 10t\)

\(t = 2.5 \text{ s}\)

(ii) To find the length of time for which P is higher than 23 m above the ground, use the equation:

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

where \(s = 20\) m (since 23 m above the ground is 20 m above the projection point), \(u = 25\) m/s, and \(a = -10\) m/s². Solving the quadratic equation:

\(20 = 25t - 5t^2\)

Rearranging gives:

\(5t^2 - 25t + 20 = 0\)

Solving this quadratic equation gives \(t = 1\) s and \(t = 4\) s. The time interval is:

\(4 - 1 = 3 \text{ s}\)

(iii) To find h such that P is higher than h m above the ground for 1 second, note that the maximum height is reached at 2.5 s. Therefore, P reaches h after 2 s. Using:

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

where \(u = 25\) m/s, \(a = -10\) m/s², and \(t = 2\) s:

\(h - 3 = 25 \times 2 - 5 \times 2^2\)

\(h - 3 = 50 - 20\)

\(h = 33 \text{ m}\)

(i) To find the maximum height, use the equation of motion:

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

where \(v = 0\) (at maximum height), \(u = 30\) m/s, and \(a = -g\) (acceleration due to gravity, \(g = 9.8\) m/s²).

\(0 = 30^2 + 2(-9.8)s\)

\(s = \frac{900}{19.6} = 45\) m

(ii) To find the time to reach 33.75 m, use:

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

\(33.75 = 30t - \frac{1}{2}gt^2\)

\(33.75 = 30t - 4.9t^2\)

\(4.9t^2 - 30t + 33.75 = 0\)

Solving this quadratic equation gives \(t = 1.5\) s (rejecting \(t = 4.5\) s as it is the second time reaching this height).

To find the speed at this time, use:

\(v = u + at\)

\(v = 30 - 9.8 \times 1.5\)

\(v = 15\) m/s

First, use the equation of motion to find \(V\):

\(V^2 = 5^2 + 2 \times g \times 7.2\)

Assuming \(g = 9.8 \text{ m s}^{-2}\), solve for \(V\):

\(V^2 = 25 + 2 \times 9.8 \times 7.2\)

\(V^2 = 25 + 141.12\)

\(V^2 = 166.12\)

\(V = \sqrt{166.12} \approx 12.89 \text{ m s}^{-1}\)

Rounding gives \(V = 13 \text{ m s}^{-1}\).

Next, find the time taken for the ball to reach the ground from A:

Using \(v = u + gt\), where \(v = 13 \text{ m s}^{-1}\), \(u = 5 \text{ m s}^{-1}\), and \(g = 9.8 \text{ m s}^{-2}\):

\(13 = 5 + 9.8t\)

\(8 = 9.8t\)

\(t = \frac{8}{9.8} \approx 0.82 \text{ s}\)

For the rebound, the initial speed is \(\frac{1}{2} V = 6.5 \text{ m s}^{-1}\):

Using \(v = u - gt\), where \(v = 0 \text{ m s}^{-1}\) at the highest point:

\(0 = 6.5 - 9.8t\)

\(9.8t = 6.5\)

\(t = \frac{6.5}{9.8} \approx 0.66 \text{ s}\)

\(Total time = 0.82 s + 0.66 s = 1.48 s\)

Rounding gives the total time as 1.45 s.