Drag force (air resistance) and acceleration

[nautikal]

Homework Statement

A baseball is thrown straight up. The drag force is proportional to v^{2}. In terms of g, what is the ball's acceleration when its speed is half its terminal speed and a) it is moving up? b) it is moving back down?

Homework Equations

\sum F=ma
mg + bv^{2}=ma - When the ball is moving up
mg - bv^{2}=ma - When the ball is moving back down

The Attempt at a Solution

We did something like this in class except we used mg - kv=ma. We went through the calculus and got an equation for v(t) and then took the derivative to get a(t). Following my notes, I did the calculus to get an equation for v(t), and then I took the derivative to get an equation for a(t) using mg - bv^{2}=ma. Below is what I got (in the attachment). I haven't done calculus like this since last May, so there may be some errors, but that's what I got.

So I got an answer for a(t) for when the ball is moving down, but I don't know what to do in order to simplify it so that it's acceleration for "when its speed is half its terminal speed".

 

[Mindscrape]

I think you are really good at nonlinear calculus or I am missing something. How did you solve

\int_0^t dt = \int_{v(0)}^{v(t)} \frac{dv}{g + \frac{b}{m} v^2}

 

[nautikal]

I think you are really good at nonlinear calculus or I am missing something. How did you solve

\int_0^t dt = \int_{v(0)}^{v(t)} \frac{dv}{g + \frac{b}{m} v^2}

I did my algebra before this step a little differently to get \frac{1}{m}dt=\frac{1}{mg-bv^{2}}dv

Then use U substitution to integrate. u=(mg-bv^{2})dv and du=-2bvdv

You need to first multiply both sides by -2bv in order to be able to use the u substitution.

So you should have \frac{-2bv}{m}dt=\frac{-2bv}{mg-bv^{2}}dv

Then plug in du and u so you have
\frac{-2bv}{m}dt=\frac{1}{u}du

Next integrate... \int_0^t \frac{-2bv}{m}dt = \int_{v(0)}^{v(t)} \frac{1}{u}du

So you get \frac{-bv^{2}t}{m}=ln\frac{mg-bv^{2}}{mg}

The rest is algebra to solve for v.
v=\sqrt{\frac{mg}{b}(1-e^\frac{-bv^{2}}{m})

 

[Mindscrape]

Yeah, the problem is that you can't directly integrate

\int_0^t \frac{-2bv}{m}dt

because there is a time dependence in v. Let me see what mathematica says.

Supposedly the solution is

v[t] = \frac{(\sqrt{g} \sqrt{m} Tan[\frac{(\sqrt{b} \sqrt{g} t)} {\sqrt{m}}])}{\sqrt{b}}

Edit:
Yep, it gave the right soln. You could probably find the integral in an integral table somewhere.

 

[nautikal]

Yeah, the problem is that you can't directly integrate

\int_0^t \frac{-2bv}{m}dt = \int_{v(0)}^{v(t)} \frac{1}{u}du

because there is a time dependence in t. Let me see what mathematica says.

Supposedly the solution is

v[t] -> \frac{(\sqrt{g} \sqrt{m} Tan[\frac{(\sqrt{b} \sqrt{g} t)} {\sqrt{m}}])}{\sqrt{b}}

*shrugs*

I just went off my notes (where we had mg-kv=ma), and in my notes we did \int_0^t \frac{-k}{m}dt = \int_{v(0)}^{v(t)} \frac{-k}{u}dv, where u=mg-kv and du=-kdv

We ended up getting v=\frac{mg}{k}(1-e^{\frac{-k}{m}t})

But this was when we had F_{air}=kv. Now we have F_{air}=bv^{2} and things are a bit harder.

I now see that we have v and t in that integral so it doesn't work. Oh well, at least my paper looks all pretty with calculus all over it :).

P.S. How do you make the LaTeX images larger? I tried the \large and \huge but it doesn't have any effect. My solution for v two posts up is difficult to read.

 

[Mindscrape]

I'm not really sure, but, if you haven't seen it already, there is a LaTeX for starters here

https://www.physicsforums.com/showthread.php?t=8997

Alternatively, just say \textrm{exp}(blah) if you don't want a cramped exponential.

Is this a physics 1 class, classical mechanics, something else? Not a trivial differential equation to solve.

 

[nautikal]

nevermind...I got it. Was overthinking this problem too much. Just needed to solve for terminal velocity and then divide by half, then plug back into the F=ma equation.

 

[pedro_infante]

what was the answer to this question?