(i) The total distance fallen is calculated by finding the area under the velocity-time graph. The graph consists of a triangle, a trapezium, and a rectangle.

Area of the triangle (first stage):

\(\frac{1}{2} \times 5 \times 50 = 125 \text{ m}\)

Area of the trapezium (second stage):

\(\frac{1}{2} \times 7 \times (8 + 50) = 203 \text{ m}\)

Area of the rectangle (third stage):

\(90 \times 8 = 720 \text{ m}\)

\(Total distance = 125 + 203 + 720 = 1048 \text{ m}\)

(ii) During the second stage, the parachutist decelerates. The acceleration \(a\) is given by:

\(a = \frac{8 - 50}{12 - 5} = -6 \text{ m/s}^2\)

Using Newton's second law, \(F = ma\), where \(m\) is the mass of the parachutist:

\(850 - F = 85 \times (-6)\)

\(850 - F = -510\)

\(F = 850 + 510 = 1360 \text{ N}\)

(i) To find the height of A above the surface of the liquid, use the area under the velocity-time graph from t = 0 to t = 0.7 s. The area represents the displacement:

Area = \(\frac{1}{2} \times 0.7 \times 7 = 2.45\) m.

Thus, the height of A above the surface of the liquid is 2.45 m.

\(To find the depth of the liquid, use the area under the graph from t = 0.7 to t = 1.2 s:\)

Area = \(\frac{1}{2} \times (7 + 5) \times (1.2 - 0.7) = 3\) m.

Thus, the depth of the liquid is 3 m.

(ii) The deceleration of the stone while it is moving in the liquid can be found using the gradient of the velocity-time graph from t = 0.7 to t = 1.2 s:

Deceleration = \(\frac{5 - 7}{1.2 - 0.7} = -4\) m s^-2.

(iii) Using Newton's second law, the net force on the stone is given by:

\(0.7 - mg = ma\)

where \(a = -4\) m s^-2 and \(g = 9.8\) m s^-2.

Rearranging gives:

\(0.7 = m(9.8 - 4)\)

\(m = \frac{0.7}{5.8} = 0.05\) kg.

To solve this problem, we need to resolve forces in two directions: radially and tangentially.

1. Radially, the forces are resolved as:

\(T \cos 25^{\circ} = 40 + R \cos 50^{\circ}\)

2. Tangentially, the forces are resolved as:

\(R \sin 50^{\circ} = T \sin 25^{\circ}\)

From these two equations, we can solve for the tension \(T\) and the normal reaction \(R\).

3. Solving the equations:

From the tangential equation:

\(R \sin 50^{\circ} = T \sin 25^{\circ}\)

From the radial equation:

\(T \cos 25^{\circ} = 40 + R \cos 50^{\circ}\)

4. Solving these equations simultaneously gives:

\(T = 72.5 \text{ N}\)

\(R = 40 \text{ N}\)

(a) To find the tensions in the strings, resolve the forces horizontally and vertically. Let the tensions in strings AC and BC be T_A and T_B respectively.

Resolving horizontally: \(T_A \times 0.6 - T_B \times 0.8 = 20\)

Resolving vertically: \(T_A \times 0.8 + T_B \times 0.6 = 10g\)

Solving these equations simultaneously:

From the horizontal resolution: \(T_A = \frac{0.8}{0.6} T_B + \frac{20}{0.6}\)

Substitute into the vertical resolution:

\(\left( \frac{0.8}{0.6} T_B + \frac{20}{0.6} \right) \times 0.8 + T_B \times 0.6 = 10g\)

Simplifying gives: \(0.87 T_A = 20 \rightarrow T_A = 76 \, \text{N}\)

Substitute back to find \(T_B\): \(T_B = 68 \, \text{N}\)

(b) For the greatest value of F, consider when \(T_B = 0\):

From the vertical resolution: \(T_A \times 0.6 = 10g \rightarrow T_A = \frac{500}{3}\)

From the horizontal resolution: \(T_A \times 0.8 = F\)

Substitute \(T_A = \frac{500}{3}\):

\(F = \frac{400}{3} \approx 133 \, \text{N}\)

(i) Resolve forces horizontally and vertically at J:

\(0.8T_A + 0.6T_R = 5.6\)

\(0.6T_A = 0.8T_R\)

Solve these simultaneous equations to find:

\(T_A = 4.48\, \text{N}\)

\(T_R = 3.36\, \text{N}\)

(ii) For limiting equilibrium, resolve vertically:

\(0.2g + F = T_R \times \cos 36.9\)

Normal force:

\(N = T_R \times \sin 36.9\)

Using \(\mu = \frac{F}{N}\):

\([0.2g + T_R \times 0.6 = T_R \times 0.8]\)

\(\mu = \frac{0.688}{2.016} = 0.341\)

(iii) For a particle of mass m:

\(0.2g + mg = \mu N + 0.8T_R\)

Substitute known values:

\(0.2g + mg = 0.341 \times 2.016 + 3.36 \times 0.8\)

Solve for m:

\(m = 0.137\, \text{or}\, 0.138\, \text{kg}\)