What is the Terminal Velocity and Drag at High Altitudes?

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AI Thread Summary
The discussion centers on calculating the air density at 90,000 feet based on a skydive scenario where an Air Force officer reached a speed of 614 mph. The equations for terminal velocity and drag are applied, but the user struggles with the calculations, particularly in determining the correct air density. Initial attempts yielded a value of 0.054, which was not validated against known densities at lower altitudes. There is also confusion regarding the desired units for the final answer, whether imperial or metric. The conversation highlights the complexities of physics calculations at high altitudes.
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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...
 
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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.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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