The vertical component of the landing speed of a parachute is 6m/s. consider parachute as an open hemisphere determine its diameter if the total weight to be carried is 1200 N . take mass density of air $1.205 kg/m^3$ , CD =1.33.

Problem Statement : - The vertical component of the landing speed of a parachute is 6m/s. consider parachute as an open hemisphere determine its diameter if the total weight to be carried is 1200 N. Take mass density of air 1.205kg/m3 , CD =1.33.

Solution : -

$Given, $

$\ \ \ Cd = 1.33$

$\ \ \ Density = 1.205 Kg / m^3$

$\ \ \ Mass = 1200 N$

$\ \ \ Velocity= 6 m/sec$

With the parachute, neglecting buoyancy force using Newton’s second law of motion:

$$ \sum F = 0$$

$Therefore, $

$\implies mg - \frac{1}{2} \rho V^2 A C_d = 0$

$\implies 1500 - \frac{1}{2} \times 1.205 \times (6)^2 \times (\frac{\pi}{4}) D^2 \times 1.4 = 0$

$\implies 1500 = 0.6025 \times 36 \times (\frac{\pi}{4}) D^2 \times 1.4 $

$\implies 1500 \times 4 = 0.6025 \times 36 \times 3.14 \times D^2 \times 1.4$

$\implies 6000 = 95.34924 \times D^2$

$\implies D^2 = 62.92656$

$\implies D = 7.932626\ m$

So, therefore it's diameter will be 7.93 m. Ans.