The force of air drag (air resistance) on an object moving through the an is F_d

Question

The force of air drag (air resistance) on an object moving through the an is F_d = C_d 1/2 rho Av^2. rho is the density of air (about 1.2 kg/m^3). A is the frontal surface area of the object. C_d depends on the shape of the object (and other factors). C_d for a flat surface is about 1. Since a parachute is an inverted surface, we expect C_4 for a parachute to be greater than 1. Websites give values between 1.3 and 1 5. We will use C_4 = 1.4. You want to design a parachute such that a person with a mass of 80 kg will fall with a terminal velocity of 3 m/sec. The parachute will have a mass of 10 kg What diameter should the chute have? Now suppose that 80 kg person is carrying 40 kg of equipment. If he uses the same parachute, what will be the terminal velocity?

Explanation / Answer

1) Here, Fd = (80 + 10) * 9.8

=> (80 + 10) * 9.8 = 1/2 * 1.4 * 1.2 * A * 32

=> A = 116.66 m2

Thus, diameter of parachute = 2 * sqrt(116.66/3.14)

                                              = 12.19 m

2)   Here, (80 + 40 + 10) * 9.8 =   1/2 * 1.4 * 1.2 * 116.66 * v2

=> v = 3.605 m/s

Thus, terminal velocity =   3.605 m/s

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