Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Calculus and Beyond Homework Help
Autonomous Diff. Equation, Free Fall Encountering Resistance
Reply to thread
Message
[QUOTE="edithc, post: 4392321, member: 477985"] [h2]Homework Statement [/h2] Given: If a body of a mass m falling from rest under the action of gravity encounters an air resistance proportional to the square of velocity, then the body's velocity t seconds in the fall satisfies v' = g - [(kv^2)/m] where k>0 and is a constant that depends on the body's aerodynamic properties and the density of the air (assume fall is too short to be affected by changes in the density). For a 110 lb skydiver (mg=110) and with time in seconds and distance in ft., a typical value of k is .005. What is the diver's terminal velocity? [h2]Homework Equations[/h2] terminal velocity = v = √((mg)/k) [h2]The Attempt at a Solution[/h2] This is a problem from the autonomous differential equation section. The first two parts of the equation were to sketch phase lines and a typical velocity curve, which was completed. This part, the third part of the question, wants the diver's terminal velocity. For the earlier sections, I found out that v=√((mg)/k) which is the equilibrium pt./terminal velocity. I expected the answer to be √(110/.005) but the back of the textbook said that it was √(160/.005). How did 110 become 160 for the mg section? Does it have something to do with a conversion that I needed to do and missed or did it involve solving with v''? Thanks in advance for any help. [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Calculus and Beyond Homework Help
Autonomous Diff. Equation, Free Fall Encountering Resistance
Back
Top