Question

A skydiver will reach a terminal velocity when the air drag equals their weight. For a skydiver with high speed and a large body, turbulence is a factor. The drag force then is approximately proportional to the square of the velocity. Taking the drag force to be FD =1/2 rhoAv2 and setting this equal to the person’s weight, find the terminal speed for a person falling "spread eagle." Find both a formula and a number for vt , with assumptions as to size.

Answer 1

Answer with Explanation:

It is given that

1) Force of Drag $F_{D}=\frac{1}{2}\rho Av^{2}$

2) Weight $W=mg$

Equating drag and weight we get

$F_{D}=W\\\\\frac{1}{2}\rho Av^{2}=mg\\\\v^{2}=\frac{2mg}{\rho A}\\\\\therefore v=\sqrt{\frac{2mg}{\rho _{air}A}}$

where,

'm' is mass of the object

'g' is acceleration due to gravity

'A' is the area of the person

$\rho _{air}$ is the density of air

Approximating the weight of a man to be 75 kg and area of man $0.8m^{2}$

Applying the values in the  formula we get

$v=\sqrt{\frac{2\times 75\times 9.81}{1.225\times 0.8}}$

$\therefore v=38.75m/s$