Without opening his parachute, a skydiver reaches a fall velocity of \(v_1 = 50\text { m/s}.\) 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 \(v_2 \) does he reach the ground?

**Details and Assumptions:**

- The air frictional force \(F_f = F_f (\rho, A, v) \) depends only on the density of the air \(\rho ,\) the cross-sectional area of the skydiver \(A, \) and the velocity \(v.\) Neglect the effect that the shape of the falling object has on the air frictional force.
- The cross-sectional area without the parachute is \(A_1 \approx 1\text { m}^2\) and that with the parachute is \(A_2 \approx 100\text{ m}^2.\)
- The density of the air is constant.

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