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?