Opening a parachute

Without opening his parachute, a skydiver reaches a fall velocity of v1=50 m/s.v_1 = 50\text { m/s}. When he does open the parachute, he's braked by additional air resistance. After a while, he finally arrives at the ground.

At what fall velocity v2v_2 does he reach the ground?

Details and Assumptions:

  • The air frictional force Ff=Ff(ρ,A,v)F_f = F_f (\rho, A, v) depends only on the density of the air ρ,\rho , the cross-sectional area of the skydiver A,A, and the velocity v.v. Neglect the effect that the shape of the falling object has on the air frictional force.
  • The cross-sectional area without the parachute is A11 m2A_1 \approx 1\text { m}^2 and that with the parachute is A2100 m2.A_2 \approx 100\text{ m}^2.
  • The density of the air is constant.

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