# Thread: Trying to formulate then solve DE [1stOrder notsure what type]

1. ## Trying to formulate then solve DE [1stOrder notsure what type]

This is about dropping a spherical object from 6m high and creating and solving a mathematical model that includes drag.

What i've done so far is.

Got that the force of drag is this

$\displaystyle \mathbf{F}_d= -{1 \over 2} \rho v^2 A C_d \mathbf{\hat v}$

Drawn a free body diagram showin gravity acting downwards[+ve] and drag opposing this[-ve]

Then using F=ma

$\displaystyle mg -{1 \over 2} \rho v^2 A C_d=ma$

(ignored the vector as im working in 1D [V hat])

Then i changed $\displaystyle -{1 \over 2} \rho v^2 A C_d$ by removing $\displaystyle v^2$ to create the constant $\displaystyle k=\frac{1}{2}\rho A C_d$

To get

$\displaystyle mg-k v^2=ma$

then using $\displaystyle a=\frac{dv}{dt}$

i get $\displaystyle mg-k v^2=m\frac{dv}{dt}$

now i'm trying to rearrange this to get my differential eqn this is where i get stuck

from the above im doing this

$\displaystyle dt(mg-k v^2)=m{dv}$

$\displaystyle dt=\frac{m}{mg-k v^2}dv$

now what do i do?

2. Originally Posted by Kevlar
This is about dropping a spherical object from 6m high and creating and solving a mathematical model that includes drag.

What i've done so far is.

Got that the force of drag is this

$\displaystyle \mathbf{F}_d= -{1 \over 2} \rho v^2 A C_d \mathbf{\hat v}$

Drawn a free body diagram showin gravity acting downwards[+ve] and drag opposing this[-ve]

Then using F=ma

$\displaystyle mg -{1 \over 2} \rho v^2 A C_d=ma$

(ignored the vector as im working in 1D [V hat])

Then i changed $\displaystyle -{1 \over 2} \rho v^2 A C_d$ by removing $\displaystyle v^2$ to create the constant $\displaystyle k=\frac{1}{2}\rho A C_d$

To get

$\displaystyle mg-k v^2=ma$

then using $\displaystyle a=\frac{dv}{dt}$

i get $\displaystyle mg-k v^2=m\frac{dv}{dt}$

now i'm trying to rearrange this to get my differential eqn this is where i get stuck

from the above im doing this

$\displaystyle dt(mg-k v^2)=m{dv}$

$\displaystyle dt=\frac{m}{mg-k v^2}dv$

now what do i do?
Integrate both sides.

3. I was thinking of doing

$\displaystyle dt = (\frac{1}{g-k v^2}dv)$

then changing the denominator to

$\displaystyle dt = \frac{1}{\sqrt g-v \sqrt k)(\sqrt g-v \sqrt k}dv$

$\displaystyle \int dt =\int \frac{1}{(\sqrt g-v \sqrt k)(\sqrt g+v \sqrt k)}dv$

what partial fractions should i get on the right hand side before integrating

$\displaystyle \int (\frac{1}{\sqrt g-v\sqrt k}+\frac{1}{\sqrt g+v\sqrt k})$
this?, and if so what does it integrate to, or can someone show me how to integrate this

would i get this?

$\displaystyle t= ln(\sqrt g-v\sqrt k)+ln(\sqrt g+v\sqrt k)$

4. Try expressing it in terms of $\displaystyle \tanh^{-1}$ since:

$\displaystyle \int\frac{1}{1-u^2}du=\tanh^{-1} u$

You can put:

$\displaystyle \int_{v_0}^{v}\frac{m}{mg-kv^2}dv$

into that form right?

Or just leave it as logs but you need factors of $\displaystyle \sqrt{k}$ in your integration above since you're letting $\displaystyle u=a+bv$ and $\displaystyle du=bdv$. Whatever works for you.