Modeling a ball thrown vertically including drag from air resistance

[MigMRF]

So I'm trying to figure out how to model a ball getting thrown vertically with the starting velocity v_0. So I've come up with a differential equation which I'm pretty sure is correct:

Where D is a constant. So far so good. My problem is solving this. This is my attempt:

And when i do this and isolate v(t) i get this:

First of all, can anyone confirm this? It should only be used between t=0 and when the graph cross the x-axis. Secondly: Is there another waay to solve this.
BTW. if anyone were curios the fuction looks like this (for m=0.05 D=0.00085 g=9.82 v_0=16)

 

[PeroK]

So I'm trying to figure out how to model a ball getting thrown vertically with the starting velocity v_0. So I've come up with a differential equation which I'm pretty sure is correct:
View attachment 254204
Where D is a constant. So far so good. My problem is solving this. This is my attempt:
View attachment 254206
View attachment 254205
And when i do this and isolate v(t) i get this:
View attachment 254207
First of all, can anyone confirm this?

That looks right (although it has $x$ instead of $t$ in there), but it can be significantly simplified by cancelling terms. Try using the parameter $\alpha = \sqrt{\frac{D}{mg}}$.

PS You could also then try $\beta = \tan(g\alpha t)$. That will sort out the inverse tangents.

PPS can you show that your solution tends to $v = v_0 - gt$ as $D \rightarrow 0$?

 

[MigMRF]

That looks right (although it has $x$ instead of $t$ in there), but it can be significantly simplified by cancelling terms. Try using the parameter $\alpha = \sqrt{\frac{D}{mg}}$.

PS You could also then try $\beta = \tan(g\alpha t)$. That will sort out the inverse tangents.

PPS can you show that your solution tends to $v = v_0 - gt$ as $D \rightarrow 0$?

Well not from the math, but it pretty sure it does looking at it from a graphical point of view.

In this example I've plottet the same things as before but with an v_0-g*t graph and they are pretty close. When you increase the mass they get closer and closer (drag goes down) while an increase in drag makes them go apart (with drag getting v=0 before nondrag ofc)

 

[PeroK]

I simplified things down to:
$$v = \frac{v_0 - \frac g \alpha \tan(\alpha t)}{1 + \frac{v_0 \alpha}{ g} \tan(\alpha t)}$$
Where $\alpha = \sqrt{\frac{Dg}{m}}$

Which might be a bit easier to work with.

 

[MigMRF]

I simplified things down to:
$$v = \frac{v_0 - \frac g \alpha \tan(\alpha t)}{1 + \frac{v_0 \alpha}{ g} \tan(\alpha t)}$$
Where $\alpha = \sqrt{\frac{Dg}{m}}$

Which might be a bit easier to work with.

Thanks! It looks better this way, that's for sure :)

 

[PeroK]

Thanks! It looks better this way, that's for sure :)

It's not hard to get from your function to that. It might be worth it to check I haven't made any mistakes.

 

[osilmag]

I will put it back up that you could also do a u sub where $u=Dv$ and $a=\sqrt{mg}$. Then use an inverse trig table.