# Upward vertical movement with air drag.

• May 19th 2011, 05:56 AM
bartroels
Upward vertical movement with air drag.
\hat{} Hi, here's the problem: "upward vertical movement with air drag"
speed = v; initial speed at t0 = v0; initial height x0 = 0
x= vertical distance travelled

force gravity Fg=-m.g
force drag Fd=-k.v^2 (k=ct describing air, shape of object,...)

F=m.a :
--> m.a=-m.g-k.v^2 (a=x'' and y=x' and devide by mass m)
--> x'' = -g - (k/m).x'^2
--> x'' + (k/m).x'^2 = -g
--> what is the function of x in time, and of v in time?
That's a DE with second order and second degree, never learned this but am eager to learn it, searched som sites about DE but didn't easily find a way to solve this...
thanks for the help!
Bart
• May 19th 2011, 06:03 AM
TheEmptySet
Quote:

Originally Posted by bartroels
\hat{} Hi, here's the problem: "upward vertical movement with air drag"
speed = v; initial speed at t0 = v0; initial height x0 = 0
x= vertical distance travelled

force gravity Fg=-m.g
force drag Fd=-k.v^2 (k=ct describing air, shape of object,...)

F=m.a :
--> m.a=-m.g-k.v^2 (a=x'' and y=x' and devide by mass m)
--> x'' = -g - (k/m).x'^2
--> x'' + (k/m).x'^2 = -g
--> what is the function of x in time, and of v in time?
That's a DE with second order and second degree, never learned this but am eager to learn it, searched som sites about DE but didn't easily find a way to solve this...
thanks for the help!
Bart

Hint:

It may be easier to solve this problem in two parts

Let the acceleration be

$a=\frac{dv}{dt}$

Now you can write this as a first order ode

$m\frac{dv}{dt}=-mg-kv^2$

This equation is first order and separable!

Then use

$v=\frac{dx}{dt}$

to finish!