First, calculate the maximum height reached by the particle. Using the equation for upward motion:

\(0 = u^2 - 2gs\)

where \(u = 9 \text{ m/s}\) and \(g = 9.8 \text{ m/s}^2\). Solving for \(s\):

\(0 = 9^2 - 2 \times 9.8 \times s\)

\(s = \frac{81}{19.6} = 4.05 \text{ m}\)

The total distance travelled is the sum of the distance upwards and downwards:

\(\text{Total distance} = 4.05 + (3.15 + 4.05) = 11.25 \text{ m}\)

Next, calculate the time taken to reach the maximum height:

\(v = u - gt\)

Setting \(v = 0\):

\(0 = 9 - 9.8t\)

\(t = \frac{9}{9.8} = 0.9 \text{ s}\)

For the downward motion, use:

\(s = \frac{1}{2} gt^2\)

where \(s = 3.15 + 4.05 = 7.2 \text{ m}\):

\(7.2 = \frac{1}{2} \times 9.8 \times t^2\)

\(t^2 = \frac{14.4}{9.8}\)

\(t = 1.2 \text{ s}\)

The total time taken is:

\(\text{Total time} = 0.9 + 1.2 = 2.1 \text{ s}\)

(i) The time for particle P to reach the ground is given by solving the equation for vertical motion:

\(17t - 5t^2 = 0\)

Solving gives \(t = 0\) or \(t = 3.4\) seconds. The time for particle Q to reach the ground is given by:

\(7(t - T) - 5(t - T)^2 = 0\)

Solving gives \(t - T = 0\) or \(t - T = 1.4\) seconds. Since both particles reach the ground at the same time, \(t = 3.4\) and \(t - T = 1.4\), so \(T = 2\) seconds.

(ii) At the instant when P is 5 m higher than Q, we have:

\(17(t + 2) - 5(t + 2)^2 - (7t - 5t^2) = 5\)

Solving gives \(t = 0.9\) or \(t = 2.9\) seconds. Using \(t = 0.9\), the velocities are:

For P: \(v_P = 17 - 10(t + 2) = 12 \text{ m s}^{-1}\)

For Q: \(v_Q = 7 - 10t = 2 \text{ m s}^{-1}\)

Both velocities are directed vertically downwards.

(i) Using the equation of motion: \(0 = u^2 - 2gs\), where \(s = 45 \text{ m}\) (40 m + 5 m) and \(g = 10 \text{ ms}^{-2}\), we have:

\(u^2 = 2 \times 10 \times 45\)

\(u = 30 \text{ ms}^{-1}\)

(ii) Using \(s = ut - \frac{1}{2}gt^2\) with \(s = 40\), \(u = 30\), and \(g = 10\):

\(40 = 30t - 5t^2\)

Solving \(5t^2 - 30t + 40 = 0\) gives \(t = 2\) and \(t = 4\).

The time above the top of the cliff is \(4 - 2 = 2 \text{ s}\).

(iii) For the second signal, the maximum height above the top of the cliff is \(\frac{1}{2}g(17 \div 4) = 21.25 \text{ m}\).

Using \(0 = V^2 - 2g(40 + 21.25)\):

\(V^2 = 2 \times 10 \times 61.25\)

\(V = 35 \text{ ms}^{-1}\)

To solve this problem, we use the equations of motion under constant acceleration due to gravity, where the acceleration is \(g = 9.8 \text{ m/s}^2\).

(i) Speed at the mid-point of OA:

The maximum height A is found using the equation: \(0 = u^2 - 2gs\), where \(u = 8 \text{ m/s}\) and \(v = 0 \text{ m/s}\) at the highest point.

\(0 = 8^2 - 2g \cdot s\)

\(s = \frac{64}{2 \times 9.8} = 3.2 \text{ m}\)

The mid-point of OA is at \(s = 1.6 \text{ m}\).

Using \(v^2 = u^2 - 2gs\) to find the speed at the mid-point:

\(v^2 = 8^2 - 2 \times 9.8 \times 1.6\)

\(v^2 = 64 - 31.36\)

\(v = \sqrt{32.64} \approx 5.66 \text{ m/s}\)

(ii) Time to reach the mid-point of OA:

Using \(v = u - gt\), where \(v = 5.66 \text{ m/s}\) at the mid-point:

\(5.66 = 8 - 9.8t\)

\(9.8t = 8 - 5.66\)

\(t = \frac{2.34}{9.8} \approx 0.234 \text{ s}\)

First, calculate the time \(t\) it takes for the object to fall 125 m under gravity. Using the formula for free fall:

\(125 = \frac{1}{2} \times 10 \times t^2\)

\(t^2 = 25\)

\(t = 5 \text{ s}\)

Next, calculate the distance \(s\) that \(P\) travels in this time. \(P\) starts with an initial speed of \(5 \, \text{m/s}\) and accelerates at \(0.8 \, \text{m/s}^2\):

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

\(s = 5 \times 5 + \frac{1}{2} \times 0.8 \times 5^2\)

\(s = 25 + 10 = 35 \text{ m}\)