written 2.4 years ago by
RakeshBhuse
• 3.2k

•
modified 2.4 years ago

Solution :
Given :
Weight of man, $W=90{~kg}$
Velocity of parachute, $u=20 {~m}/{~s}$
Coefficient 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,
$\therefore D=\sqrt{8.9946}=2.999{~m}$