Assume$ Cd = 0.5, \rho = 1.25 kg/m^3$.

A man weighing 90kg descends from an aeroplane with parachute against the resistance to air. The velocity with which the parachute which is hemispeherical in shape comes down in 20m/s. Find the diameter of parachute. Assume$ Cd = 0.5, \rho = 1.25 kg/m^3$.

1 Answer

Solution :

Given :

Weight of man, $W=90{~kg}$

Velocity of parachute, $u=20 {~m}/{~s}$

Co-efficient of drag, $C_{D}=0.5$

Density of air, $\rho=1.25 {~kg}/{~m}^{3}$

Let the diameter of parachute $=D$

Drag $(F_D)$

$\begin{aligned} \quad F_{D} &=90{~kg}=90 \times 9.81\\ &=883{~N}\end{aligned}$

We know that

$ \begin{aligned} \therefore F_{D} &=C_{D} \times A \times \frac{\rho u^{2}}{2}\\ \therefore 883 &=0.5 \times \frac{\pi}{4} D^{2} \times \frac{1.25 \times 20^{2}}{2} \\ \therefore D^{2} &=\frac{883 \times 4 \times 2.0}{0.5 \times \pi \times 1.25 \times 20 \times 20}\\ &=8.9946 \end{aligned} $

Diameter of parachute, $\therefore D=\sqrt{8.9946}=2.999{~m}$

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