(i) To find the greatest height, use the equation of motion: \(v^2 = u^2 + 2as\). Here, \(v = 0\) at the greatest height, \(u = 6\) m/s, \(a = -g = -9.8\) m/s², and \(s = 0.5\) m.

First, find the initial velocity \(u\) when the particle is at 0.5 m:

\(6^2 = u^2 - 2 \times 9.8 \times 0.5\)

\(36 = u^2 - 9.8\)

\(u^2 = 46\)

Now, find the total height:

\(0 = 46 - 2 \times 9.8 \times s\)

\(2 \times 9.8 \times s = 46\)

\(s = \frac{46}{19.6} = 2.3\) m

(ii) To find the time to return to O, use \(s = ut + 0.5gt^2\) with \(s = 2.3\) m, \(u = 0\), and \(g = 9.8\) m/s²:

\(2.3 = 0 + 0.5 \times 9.8 \times t^2\)

\(t^2 = \frac{2.3}{4.9}\)

\(t = \sqrt{\frac{2.3}{4.9}} = 0.678\) s

Total time for the round trip is \(2 \times 0.678 = 1.36\) s

(a) To find the speed of P when it is 10 m above the ground, use the equation of motion:

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

where \(u = 15 \text{ m s}^{-1}\), \(a = -g = -9.8 \text{ m s}^{-2}\), and \(s = 10 \text{ m}\).

Substitute the values:

\(v^2 = 15^2 - 2 \times 9.8 \times 10\)

\(v^2 = 225 - 196\)

\(v^2 = 29\)

\(v = \sqrt{29} \approx 5 \text{ m s}^{-1}\)

(b) For the collision, use the equations of motion for both particles:

For P: \(s_P = 15t - \frac{1}{2}gt^2\)

For Q: \(s_Q = 18 - \frac{1}{2}gt^2\)

Set \(s_P + s_Q = 18\) and solve for \(t\):

\(15t - \frac{1}{2}gt^2 + 18 - \frac{1}{2}gt^2 = 18\)

\(15t = gt^2\)

\(t = \frac{15}{g}\)

Substitute \(t\) back into \(s_P\):

\(s_P = 15 \times \frac{15}{g} - \frac{1}{2}g \left(\frac{15}{g}\right)^2\)

\(s_P = 10.8 \text{ m}\)

To solve for V and H, we use the equations of motion under gravity. Assume acceleration due to gravity is \(g\).

(i) Finding \(V\):

Using the equation \(v^2 = u^2 + 2gs\), where \(u\) is the initial velocity, \(v\) is the final velocity, and \(s\) is the displacement:

For the section from 35 m above the ground to the ground:

\(V^2 = (V - 10)^2 + 2g \times 35\)

Expanding and simplifying:

\(V^2 = V^2 - 20V + 100 + 70g\)

\(20V = 100 + 70g\)

Assuming \(g = 10 \text{ m s}^{-2}\):

\(20V = 100 + 700\)

\(20V = 800\)

\(V = 40\)

(ii) Finding \(H\):

Using the equation \(v^2 = u^2 + 2gs\) for the entire fall from H to the ground:

\(40^2 = 0^2 + 2gH\)

\(1600 = 20H\)

\(H = 80\)

(i) To find the time after projection at which P hits the ground, we use the equation of motion:

\(v = u - gt\)

where \(v = -11\) m/s (final velocity when hitting the ground), \(u = 11\) m/s (initial velocity), and \(g = 10\) m/s² (acceleration due to gravity).

Substitute the values:

\(-11 = 11 - 10t\)

Solve for \(t\):

\(-11 = 11 - 10t\)

\(-22 = -10t\)

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

(ii) To find the values of \(h\) and \(V\):

For particle Q, using the equation:

\(h = 0 + \frac{1}{2} g t^2\)

Substitute \(g = 10\) m/s² and \(t = 2.2\) s:

\(h = \frac{1}{2} \times 10 \times (2.2)^2 = 24.2 \text{ m}\)

For the speed \(V\) of Q when it hits the ground:

\(V = 0 + gt\)

Substitute \(g = 10\) m/s² and \(t = 2.2\) s:

\(V = 10 \times 2.2 = 22 \text{ m/s}\)

(i) First, calculate the speed of P just before entering the liquid using the equation:

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

where \(u = 0\), \(a = g = 9.8 \text{ m s}^{-2}\), and \(s = 7.2 \text{ m}\).

\(v^2 = 2 \times 9.8 \times 7.2\)

\(v = 12 \text{ m s}^{-1}\)

Now, use the equation \(v^2 = u^2 + 2as\) for the motion in the liquid:

\(6^2 = 12^2 + 2a \times 0.8\)

\(a = -67.5 \text{ m s}^{-2}\)

Using Newton's second law, \(0.2g - R = 0.2 \times 67.5\)

\(R = 15.5 \text{ N}\)

(ii) For the upward motion, use \(s = \frac{1}{2}at^2\) to find \(a\):

\(3.6 = \frac{1}{2}a \times 4^2\)

\(a = 0.45 \text{ m s}^{-2}\)

Using Newton's second law, \(T - R - 0.2g = 0.2 \times 0.45\)

\(T = 17.6 \text{ N}\)