A man descends to the ground from an aircraft with the help of a hemispherical parachute of 3 m diameter against the resistance of air. If the weight of the man is 690 N, find the velocity of parachute with which it comes down. The density of air is 1.2 \mathrm{~kg} / \mathrm{m}^3 \text { and. } C_D=0.4 for the hemisphere.

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*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}

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