An 80.0-kg skydiver jumps out of a balloon at an altitude of 1,000 m and opens his parachute at an altitude of 200 m. A. Assuming the total friction (resistive) force on the skydiver is constant at 50.0 N with the parachute closed and constant at 3,600 N with the parachute open, find the speed of the skydiver when he lands on the ground. B. At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 m/s

Explanation:

m = Mass of skydiver = 80 kg

= Height for which the parachute is closed = 1000-200 = 800 m

= Height for which the parachute is open = 200 m

= Resistive force when parachute is closed = 50 N

= Resistive force when parachute is open = 3600 N

v = Velocity of skydiver on the ground

g = Acceleration due to gravity =

h = Height from which the skydiver jumps = 1000 m

The energy balance of the system will be

The velocity fo the skydiver when he lands will be

x = Height where the person opens the parachute

v = 5 m/s

The height at which the parachute is to be opened is

d is the correct answer