This problem involves energy conservation and dynamics in a pulley system. Let's analyze the situation carefully:

Part 1: Speed at which the crate hits the ground

We will use the principle of energy conservation to solve this problem. Since there is no friction or air resistance, the total mechanical energy is conserved.

1. Before the crate falls:
   • The crate is initially 14.2 m above the ground.
   • The initial potential energy of the crate is converted into the kinetic energy of both the crate and the crewman when the system starts moving.
   • We can set up the equation for mechanical energy conservation between the initial state (when the crate is at 14.2 m above the ground) and the final state (when the crate hits the ground).

Initial energy:

• Crate's potential energy: $U_{\text{crate}} = m_{\text{crate}} \cdot g \cdot h_{\text{crate}}$
• Crewman's potential energy: $U_{\text{crewman}} = 0$ (since he's at the same level throughout)
• Initial kinetic energy: $K_{\text{initial}} = 0$ (since both are initially at rest after slipping)

Where:

• $m_{\text{crate}} = 6.95 \, \text{kg}$ (mass of the crate)
• $m_{\text{crewman}} = 50.3 \, \text{kg}$ (mass of the crewman)
• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity)
• $h_{\text{crate}} = 14.2 \, \text{m}$ (initial height of the crate)

Final energy:

• When the crate reaches the ground, both the crate and the crewman are moving with the same speed $v$ (since they are connected by the rope).
• The total kinetic energy at the final state is: $K_{\text{final}} = \frac{1}{2} m_{\text{crate}} v^2 + \frac{1}{2} m_{\text{crewman}} v^2 = \frac{1}{2} (m_{\text{crate}} + m_{\text{crewman}}) v^2$

Using the conservation of mechanical energy: $m_{\text{crate}} g h_{\text{crate}} = \frac{1}{2} (m_{\text{crate}} + m_{\text{crewman}}) v^2$

Substitute the values: $6.95 \cdot 9.8 \cdot 14.2 = \frac{1}{2} (6.95 + 50.3) v^2$

Now solve for $v$. First calculate the left-hand side: $6.95 \cdot 9.8 \cdot 14.2 = 967.286$

So: $967.286 = \frac{1}{2} \cdot 57.25 \cdot v^2$

Multiply both sides by 2: $1934.572 = 57.25 \cdot v^2$

Solve for $v^2$: $v^2 = \frac{1934.572}{57.25} = 33.8$

Now take the square root to find $v$: $v = \sqrt{33.8} \approx 5.81 \, \text{m/s}$

Thus, the crate hits the ground at a speed of approximately 5.81 m/s.

Part 2: Speed at which the crewman hits the bottom of the ravine

Since the crate pulls the crewman as it falls, the crewman will also have the same final speed when he is pulled over the edge and falls to the bottom. So, the crewman will also hit the bottom of the ravine at the same speed as the crate: 5.81 m/s.

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Summary of answers:

1. The crate will hit the ground at a speed of 5.81 m/s.
2. The crewman will hit the bottom of the ravine at a speed of 5.81 m/s.

Would you like further clarification or more detailed steps for any part? Here are a few additional questions that could be explored:

1. How would the presence of friction change the speed at which the crate hits the ground?
2. What is the total kinetic energy of the system just before the crate hits the ground?
3. How long does it take for the crate to hit the ground after the slip?
4. How much work does gravity do on the crate during its fall?
5. What is the tension in the rope during the crate's fall?

Tip: Whenever energy is conserved, focus on potential and kinetic energy transformations to simplify calculations!