# Terminal velocity/drag

• bigboss
In summary, to determine the density of air at 90000 ft, we can use the equations for terminal velocity at two altitudes and solve for density. With the given information, we can estimate the density to be approximately 0.054 kg/m^3 or 0.074887 lbm/ft^3 at 90000 ft. However, the units for the answer were not specified, so the exact value may vary depending on whether the units are in metric or imperial.

## Homework Statement

The fastest recorded skydive was by an Air Force officer who jumped from a helium balloon at an elevation of 103000 ft, three times higher than airliners fly. Because the density of air is so small at these altitudes, he reached a speed of 614 mph at an elevation of 90000 ft, then gradually slowed as the air became more dense. Assume that he fell in the spread-eagle position and that his low-altitude terminal speed is 125 mph. Use this information to determine the density of air at 90000 ft.

## Homework Equations

V=sqrt(4W/pA) w= weight, p=coefficent, a = area ... D=.25pv^2A

## The Attempt at a Solution

no idea what to do...

Consider the two equations at the two altitudes for terminal velocity.

Fdrag = m*g = 1/2*Cd*p*A*v2

Since the weight is the same ...

1/2*Cd*p90*A*V902 = 1/2*Cd*po*A*Vo2

p90*V902 = po*Vo2

i got .054, however that did not work... i used 1.29 as the p for the low altitude

bigboss said:
i got .054, however that did not work... i used 1.29 as the p for the low altitude

What units do they want the answer in?

Imperial or metric?

Wikipedia said:
At 20 °C and 101.325 kPa, dry air has a density of 1.2041 kg/m3.
At 70 °F and 14.696 psia, dry air has a density of 0.074887 lbm/ft3.

.054 kg/m^3.