Final answer:
The terminal velocity of John Kruger would be 38.83 m/s.
Step-by-step explanation:
The terminal velocity of a skydiver can be calculated using the equation:
v = sqrt((2mg)/(ρAC))
where v is the terminal velocity, m is the mass of the skydiver, g is the acceleration due to gravity, ρ is the density of the air, A is the frontal area, and C is the drag coefficient.
In this case, the frontal area of John Kruger is 1 m² and the area of the closed chute is 0.5 m². Assuming a drag coefficient of 1 and a density of air of 1.225 kg/m³, we can calculate:
v = sqrt((2 * 75 * 9.8) / (1.225 * 1 * 1)) = 38.83 m/s
Therefore, the terminal velocity of John Kruger would be 38.83 m/s.