# 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

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

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

Then using F=ma

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

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

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

To get

$mg-k v^2=ma$

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

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

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

from the above im doing this

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

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

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

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

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

Then using F=ma

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

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

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

To get

$mg-k v^2=ma$

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

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

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

from the above im doing this

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

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

now what do i do?
Integrate both sides.

3. I was thinking of doing

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

then changing the denominator to

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

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

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

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

would i get this?

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

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

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

You can put:

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

into that form right?

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