(i) To find the greatest height above the ground, use the equation of motion:

\(0 = 5.2^2 - 2 imes 10.4 imes s_1\)

\(s_1 = \frac{5.2^2}{2 imes 10.4} = 1.3 ext{ m}\)

The total height above the ground is:

\(6.2 + 1.3 = 7.5 ext{ m}\)

(ii) To find the speed with which P reaches the ground, use the equation:

\(v^2 = 2 imes 9.6 imes 7.5\)

\(v = \sqrt{2 imes 9.6 imes 7.5} = 12 ext{ m/s}\)

(iii) To find the total work done by the air resistance, calculate the energy changes:

\(Potential energy loss = 0.6 imes 9.8 imes 6.2 = 37.2 ext{ J}\)

\(Initial total energy = 0.6 imes 9.8 imes 6.2 + \frac{1}{2} imes 0.6 imes 5.2^2 = 45.312 ext{ J}\)

\(Energy loss upward = \frac{1}{2} imes 0.6 imes 5.2^2 - 0.6 imes 9.8 imes 1.3 = 0.312 ext{ J}\)

\(Kinetic energy gain = \frac{1}{2} imes 0.6 imes 12^2 = 43.2 ext{ J}\)

\(Energy loss downward = -\frac{1}{2} imes 0.6 imes 12^2 + 0.6 imes 9.8 imes 7.5 = -1.8 ext{ J}\)

\(Total work done = 37.2 - 35.088 + 45.312 - 43.2 = 2.11 ext{ J}\)

The work done \((WD)\) by the force is given by the formula:

\(WD = Fd \cos \theta\)

where \(F = 30 \text{ N}\) is the force, \(d\) is the distance moved, and \(\theta\) is the angle of the force above the horizontal.

Since the block moves at a constant speed of 1.5 m/s for 20 seconds, the distance \(d\) is:

\(d = 1.5 \times 20 = 30 \text{ m}\)

Substituting the given values into the work done formula:

\(720 = 30 \times 30 \cos \theta\)

Simplifying gives:

\(720 = 900 \cos \theta\)

\(\cos \theta = \frac{720}{900} = 0.8\)

Therefore, \(\theta = \cos^{-1}(0.8) \approx 36.9^\circ\).

(i) The work done by the pulling force is calculated using the formula for work: \(W = F \cdot d \cdot \cos(\theta)\), where \(F = 25 \text{ N}\), \(d = 2 \text{ m}\), and \(\theta = 15^\circ\). Thus, \(W = 25 \times 2 \times \cos(15^\circ) = 48.3 \text{ J}\).

(ii) To find the normal component of the contact force, resolve the forces vertically. The equation is \(N + 25 \sin(15^\circ) = 3 \times 10\). Solving for \(N\), we get \(N = 30 - 25 \sin(15^\circ) = 23.5 \text{ N}\).

The work done by the cyclist is given by the formula:

\(\text{Work done by cyclist} = \text{Force} \times \text{Distance} = 100 \times 50 = 5000 \text{ J}\)

The net work done is the work done by the cyclist minus the work done against the resistance force:

\(5000 - 3560 = 1440 \text{ J}\)

This net work done is equal to the change in kinetic energy of the cyclist and bicycle:

\(\frac{1}{2} m v^2 = 1440\)

Substitute \(v = 6 \text{ m s}^{-1}\):

\(\frac{1}{2} m (6)^2 = 1440\)

\(18m = 1440\)

\(m = \frac{1440}{18} = 80 \text{ kg}\)

Therefore, the total mass of the cyclist and bicycle is 80 kg.

To find the work done by the winch, we first calculate the force exerted by the winch. The force required to pull the load up the plane includes overcoming both the gravitational component along the plane and the constant resistance.

The gravitational force component along the plane is given by:

\(F_g = mg \sin \theta = 50g \sin 60^{\circ}\)

Substituting the values, we have:

\(F_g = 50 \times 9.8 \times \sin 60^{\circ} = 50 \times 9.8 \times 0.866 = 433.0 \text{ N}\)

The total force exerted by the winch is the sum of this gravitational component and the constant resistance:

\(F = F_g + 100 = 433.0 + 100 = 533.0 \text{ N}\)

The work done by the winch is the product of this force and the distance moved along the plane:

\(\text{Work done} = F \times d = 533.0 \times 5 = 2670 \text{ J}\)

Thus, the work done by the winch is 2670 J.