# Opening a parachute

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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