Question 15.11: A man descends to the ground from an aircraft with the help ......

Given data:

Diameter of parachute                          D = 3 m

Weight of man                                        W = 690 N
Coefficient of drag                                  C_D=0.4

Density of air 1.2 kg/m³

Projected area is                                    A=\frac{\pi}{4} D^2=\frac{\pi}{4}(3)^2=7.068 \mathrm{~m}^2

The total weight of the man is balanced by the drag force. Thus, the drag force is

F_D=W=690 \mathrm{~N}

Let the velocity of parachute be U.

From Eq. (15.14), we have

F_{D}=C_{D}A\frac{1}{2}\rho U_{∞}^{2}  (15.14)

F_D=C_D A \frac{1}{2} \rho U^2

or                                  U=\sqrt{\frac{F_D \times 2}{C_D A \rho}}=\sqrt{\frac{690 \times 2}{0.4 \times 7.068 \times 1.2}}=20.168 \mathrm{~m} / \mathrm{s}