The skydiver's downward velocity when the parachute opens can be calculated using the equation of motion:
vf = vi + g * t
where vf is the final velocity, vi is the initial velocity (0 m/s, since the skydiver is in free fall), g is the acceleration due to gravity (9.8 m/s^2), and t is the time (9.0 s).
vf = 0 + 9.8 m/s^2 * 9.0 s = 88.2 m/s
So, the skydiver's downward velocity when the parachute opens is 88.2 m/s.
The distance the skydiver falls during the 9.0 seconds can be calculated using the equation:
d = vi * t + 0.5 * g * t^2
where d is the distance, vi is the initial velocity (0 m/s), g is the acceleration due to gravity (9.8 m/s^2), and t is the time (9.0 s).
d = 0 * 9.0 s + 0.5 * 9.8 m/s^2 * 9.0 s^2 = 405 m
So, the skydiver is 405 meters below the helicopter when the parachute opens