Answer with Explanation:
It is given that
1) Force of Drag [tex]F_{D}=\frac{1}{2}\rho Av^{2}[/tex]
2) Weight [tex]W=mg[/tex]
Equating drag and weight we get
[tex]F_{D}=W\\\\\frac{1}{2}\rho Av^{2}=mg\\\\v^{2}=\frac{2mg}{\rho A}\\\\\therefore v=\sqrt{\frac{2mg}{\rho _{air}A}}[/tex]
where,
'm' is mass of the object
'g' is acceleration due to gravity
'A' is the area of the person
[tex]\rho _{air}[/tex] is the density of air
Approximating the weight of a man to be 75 kg and area of man [tex]0.8m^{2}[/tex]
Applying the values in the formula we get
[tex]v=\sqrt{\frac{2\times 75\times 9.81}{1.225\times 0.8}}[/tex]
[tex]\therefore v=38.75m/s[/tex]
