# Air drag and kinematics problem

95 views

A body of mass m thrown straight up with velocity $v_0$. Find the velocity v' with which the body comes down if air drag equals to $kv^2$ , where k is a constant and v is velocity of the body.

I thought :
$mg- kv^2 =m v dv/ dx$
$mg- kv^2= m vdv / dx$
$mg=kv^2$
$mg/k = v^2$
$v_t = \sqrt{mg/k}$
where $v_t$ is terminal velocity but answer given is
$v= v_0 / \sqrt{1+k{v_0}^2 / mg}$
Where I am thinking wrong ?

edited Jun 18, 2017
Please show your attempt. Write the equation of motion and try to solve it. Time is not required so you can use the substitution $a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$ to find $v(x)$.
I have edited the question :)
You have not included gravity. The equations of motion will be different for ascent and descent, because resistance changes direction. Height must be the same because the  body comes back to its launch point.
The problem does not ask for the terminal velocity. You cannot assume that terminal velocity is reached. You have to solve the differential equations for upward and downward motions.
Okay I have realized my mistake :)

1 vote

The equations of motion for the upward and downward phases are
$m v\frac{dv}{dx}=-mg-kv^2$
$m v\frac{dv}{dx}=-mg+kv^2$
in which I am taking the upward direction as +ve, and I have applied the transformation $\dot v=\frac{dv}{dt}=v\frac{dv}{dx}$. Using $c=k/mg$ these equations can be written as
$2cv\frac{dv}{dx}=-2cg(1+cv^2)$
$-2cv\frac{dv}{dx}=2cg(1-cv^2)$.

Integrate with the limits $(x,v) = (0,v_0)\to (h,0)$ and $(h,0) \to (0,-v)$. We get
$\ln (1+cv_0^2) - \ln (1)= -2cg(0-h)=2cgh$
$1+cv_0^2 = e^{2cgh}$
and
$\ln (1-cv^2) -\ln (1)= 2cg(0-h)=-2cgh$
$1-cv^2 = e^{-2cgh}$

Multiply the two solutions to eliminate $h$ :
$(1+cv_0^2)(1-cv^2) = 1$
$cv^2 =1-\frac{1}{1+cv_0^2}=\frac{cv_0^2}{1+cv_0^2}$
$v=\frac{v_0}{\sqrt{1+cv_0^2}}$.

answered Jun 19, 2017 by (27,948 points)
selected Jun 19, 2017
Today I'm diving back to this question(fun part is I'm in college now;)) I'm learning to solve differential equations now :)
a Long time though!
Thanks Sammy Gerbil :) You do great work