What is the Terminal Velocity and Drag at High Altitudes?

  • Thread starter Thread starter bigboss
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around determining the density of air at high altitudes, specifically at 90,000 feet, based on the context of terminal velocity during a skydive. The problem involves understanding the relationship between drag force, terminal velocity, and air density at varying altitudes.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the equations governing terminal velocity and drag force, considering the weight of the object and the changing air density at different altitudes. Some participants question the validity of their calculations and the units required for the answer.

Discussion Status

There is an ongoing exploration of the calculations related to air density, with some participants providing numerical attempts. However, there is no explicit consensus on the correct value or approach, and questions about unit preferences remain unresolved.

Contextual Notes

Participants note the importance of using the correct air density values for low altitude and the potential confusion regarding the units of measurement (imperial vs. metric). There is also mention of standard conditions for air density at specific temperatures and pressures.

bigboss
Messages
28
Reaction score
0

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=coefficient, a = area ... D=.25pv^2A

The Attempt at a Solution



no idea what to do...
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
5K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
12K
Replies
7
Views
16K
  • · Replies 3 ·
Replies
3
Views
20K