Finding the equation for Velocity with Drag

[Theorγ]

Homework Statement

Given a situation where a ball of mass m is dropped off at a height of h, find the equation that would give the velocity of the ball with respect to time given that gravity and drag are intrinsic factors.

Homework Equations

$$F_{g} = m g$$
$$F_{D} = \frac{1}{2}pCAv^2$$

The Attempt at a Solution

$$F_{net}= F_{g} - F_{D}$$
$$F_{net} = m g - \frac{1}{2}pCAv^2$$
$$a_{net} = g - \frac{1}{2m}pCAv^2$$
$$\frac{dv}{dt} = g - \frac{1}{2m}pCAv^2$$
$$\int \frac{1}{g - \frac{1}{2m}pCAv^2} dv = \int dt$$

Now I'm stuck; what in the world would you do to integrate the left side? I tried relating $$\int\frac{1}{1 - x^2} dx = \int\frac{-1}{(x-1)(x+1)} dx = \frac{1}{2}ln(x + 1) - \frac{1}{2}ln(x - 1) + C$$ to the left side and that integrates out by factoring and using partial fractions. However, I can't find any way to easily factor, if the left side is even factorable to begin with...

_________________________
$$\int \frac{1}{(\sqrt{g})^2 - (\sqrt{\frac{1}{2m}pCA}v)^2} dv = t + Constant$$
$$\int \frac{1}{(\sqrt{g} + (\sqrt{\frac{1}{2m}pCA}v))(\sqrt{g} - (\sqrt{\frac{1}{2m}pCA}v))} dv = t + Constant$$
$$\int \frac{\alpha}{(\sqrt{g} + (\sqrt{\frac{1}{2m}pCA}v))} dv + \int \frac{\beta}{(\sqrt{g} - (\sqrt{\frac{1}{2m}pCA}v))} dv = t + Constant$$
$$\int \frac{\frac{1}{2\sqrt{g}}}{(\sqrt{g} + (\sqrt{\frac{1}{2m}pCA}v))} dv + \int \frac{\frac{1}{2\sqrt{g}}}{(\sqrt{g} - (\sqrt{\frac{1}{2m}pCA}v))} dv = t + Constant$$
$$u_{1} = \sqrt{g} + \sqrt{\frac{1}{2m}pCA}v$$
$$\frac{du_{1}}{dv_{1}} = \sqrt{\frac{1}{2m}pCA}$$
$$u_{2} = \sqrt{g} - \sqrt{\frac{1}{2m}pCA}v$$
$$\frac{du_{2}}{dv_{2}} = -\sqrt{\frac{1}{2m}pCA}$$
$$\int \frac{\frac{1}{2\sqrt{g}}}{u_{1}} \frac{du}{\sqrt{\frac{1}{2m}pCA}} - \int \frac{\frac{1}{2\sqrt{g}}}{u_{2}} \frac{du}{\sqrt{\frac{1}{2m}pCA}} = t + Constant$$
$$\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} ln(\sqrt{g} + \sqrt{\frac{1}{2m}pCA}v) - \frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} ln(\sqrt{g} - \sqrt{\frac{1}{2m}pCA}v) = t + Constant$$

 

[da_nang]

You're on the right track.
If you substitute ½pCA/m for e.g. k, you can easily use the identity a² - b² = (a-b)(a+b), from which you can integrate with partial fractions.

It's going to get messy, though.

 

[Filip Larsen]

In case you don't have any references to check your result against you may find https://www.physicsforums.com/showthread.php?t=393048 useful.

 

[Theorγ]

So now that I've solved the velocity function, is there a way to integrate to find the position function of time?

 

[da_nang]

Well, v = ds/dt, so if you simplify the formula you can integrate it directly.

You're probably going to need to pull out some hyperbolic trigonometry, though.

Although you can do it without hyperbolic trigonometry, too.

 

[Theorγ]

Well I managed to manipulate and simplify the equation into this, but I haven't learned how to use hyperbolic functions, so what do I do?

$$a = \frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}}$$
$$b = \sqrt{\frac{1}{2m}pCA}$$
$$\frac{a}{\sqrt{g}}\frac{ds}{dt} = \frac{e^{bt} - 1}{e^{bt} + 1}$$
$$\int\frac{a}{\sqrt{g}} ds= \int\frac{e^{bt} - 1}{e^{bt} + 1} dt$$

____________
EDIT: Okay so I read up briefly on how hyperbolic functions work and this is what I've come up with:
$$\int\frac{a}{\sqrt{g}} ds= \int\tanh(\frac{bt}{2}) dt$$
$$\frac{a}{\sqrt{g}}s =\frac{2}{b} ln(cosh(\frac{bt}{2})) + Constant$$

 

[Theorγ]

Can someone confirm if I manipulated the hyperbolic function correctly and if I integrated correctly?

 

[da_nang]

Your integration looks correct. However,

$$\frac{a}{\sqrt{g}}\frac{ds}{dt} = \frac{e^{bt} - 1}{e^{bt} + 1}$$

there's an error in your simplification. You've forgotten there's an $$\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}}$$ before the natural logarithms.

I.e.

$$\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}}\times ln\left(\frac{|\sqrt{g}+v\sqrt{\frac{1}{2m}pCA}|}{|\sqrt{g}-v\sqrt{\frac{1}{2m}pCA}|}\right)=t+D$$

 

[Theorγ]

So it should be solved like this?:

$$\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} \times ln\left(\frac{|\sqrt{g}+v\sqrt{\frac{1}{2m}pCA}|}{ |\sqrt{g}-v\sqrt{\frac{1}{2m}pCA}|}\right)=t+D$$

Because the object is at rest at time 0, then D is:
_____________________________________________________________
$$\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} \times ln\left(\frac{|\sqrt{g}+0\sqrt{\frac{1}{2m}pCA}|}{ |\sqrt{g}-0\sqrt{\frac{1}{2m}pCA}|}\right) = 0+D$$
$$\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} \times ln\left(\frac{|\sqrt{g}|}{ |\sqrt{g}|}\right) = D$$
$$\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} \times ln (1) = D = 0$$
_____________________________________________________________
Using these variables as the following expressions:
$$\alpha = \frac{1}{2\sqrt{g}}$$
$$\beta = \sqrt{\frac{1}{2m}pCA}$$
_____________________________________________________________
The substitution:
$$\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} \times ln\left(\frac{|\sqrt{g}+v\sqrt{\frac{1}{2m}pCA}|}{ |\sqrt{g}-v\sqrt{\frac{1}{2m}pCA}|}\right)=t$$
$$\frac{\alpha}{\beta} \times ln\left(\frac{|\sqrt{g}+v\beta|}{ |\sqrt{g}-v\beta|}\right)=t$$
$$ln\left(\frac{|\sqrt{g}+v\beta|}{ |\sqrt{g}-v\beta|}\right)=\frac{\beta}{\alpha}t$$
$$e^{ln\left(\frac{|\sqrt{g}+v\beta|}{ |\sqrt{g}-v\beta|}\right)}=e^{\frac{\beta}{\alpha}t}$$
$$\frac{\sqrt{g}+v\beta}{\sqrt{g}-v\beta}=e^{\frac{\beta}{\alpha}t}$$

 

[Theorγ]

Continued from above:
$$\sqrt{g}+v\beta=e^{\frac{\beta}{\alpha}t} \times (\sqrt{g}-v\beta)$$
$$\sqrt{g}+v\beta=e^{\frac{\beta}{\alpha}t}\sqrt{g}-e^{\frac{\beta}{\alpha}t}v\beta$$
$$e^{\frac{\beta}{\alpha}t}v\beta+v\beta=e^{\frac{ \beta }{\alpha}t} \sqrt{g}-\sqrt{g}$$
$$v\beta \times (e^{\frac{\beta}{\alpha}t}+1)=\sqrt{g} \times (e^{\frac{\beta}{\alpha}t}-1)$$
$$\frac{\beta}{\sqrt{g}}v=\frac{(e^{\frac{\beta}{ \alpha }t}-1)}{(e^{\frac{\beta}{\alpha}t}+1)}$$
$$\frac{\beta}{\sqrt{g}}\frac{dx}{dt}=\frac{(e^{ \frac {\beta}{\alpha}t}-1)}{(e^{\frac{\beta}{\alpha}t}+1)}$$
$$\int\frac{\beta}{\sqrt{g}}dx=\int\frac{(e^{\frac{ \beta }{ \alpha }t}-1)}{(e^{\frac{\beta}{\alpha}t}+1)}dt$$
____________________________________________
Hyperbolic Substitution:
$$\frac{e^{2t}-1}{e^{2t}+1}=tanh(t)$$
$$\frac{e^{\frac{ \beta }{ \alpha }t}-1}{e^{\frac{\beta}{\alpha}t}+1}=tanh(\frac{\beta}{2\alpha}t)$$
$$\int\frac{\beta}{\sqrt{g}}dx=\int\tanh(\frac{\beta}{2\alpha}t)dt$$
$$\frac{\beta}{\sqrt{g}}x=\frac{2 \alpha }{\beta} \times ln|cosh(\frac{\beta}{2 \alpha }t)| + C$$
$$cosh(t)=\frac{e^{2t}+1}{2e^t}$$
$$cosh(\frac{\beta}{2 \alpha }t)=\frac{e^{\frac{\beta}{ \alpha }t}+1}{2e^{\frac{\beta}{2 \alpha }t}}$$
$$\frac{\beta}{\sqrt{g}}x=\frac{2 \alpha }{\beta} \times ln|\frac{e^{\frac{\beta}{ \alpha }t}+1}{2e^{\frac{\beta}{2 \alpha }t}}| + C$$

 

[da_nang]

Looks fine to me.

 

[Theorγ]

Because the object is at rest at time 0, then the object has moved a distance of 0:
$$\frac{\beta}{\sqrt{g}}x=\frac{2 \alpha }{\beta} \times ln|\frac{e^{\frac{\beta}{ \alpha }t}+1}{2e^{\frac{\beta}{2 \alpha }t}}| + C$$
$$\frac{\beta}{\sqrt{g}}0=\frac{2 \alpha }{\beta} \times ln|\frac{e^{\frac{\beta}{ \alpha }0}+1}{2e^{\frac{\beta}{2 \alpha }0}}| + C$$
$$0=\frac{2 \alpha }{\beta} \times ln|\frac{2}{2}| + C$$
$$0=0 + C$$
$$C = 0$$
______________________________________________________
Completing the equation:
$$\frac{\beta}{\sqrt{g}}x=\frac{2 \alpha }{\beta} \times ln|\frac{e^{\frac{\beta}{ \alpha }t}+1}{2e^{\frac{\beta}{2 \alpha }t}}|$$
$$x=\frac{2 \alpha \sqrt{g}}{\beta^2} \times ln|\frac{e^{\frac{\beta}{ \alpha }t}+1}{2e^{\frac{\beta}{2 \alpha }t}}|$$
____________________________________
Previous Variable Denotation:
$$\alpha = \frac{1}{2\sqrt{g}}$$
$$\beta = \sqrt{\frac{1}{2m}pCA}$$
____________________________________
Simplification:
$$x=\frac{2 \alpha \sqrt{g}}{\beta^2} \times ln|\frac{e^{\frac{\beta}{ \alpha }t}+1}{2e^{\frac{\beta}{2 \alpha }t}}|$$
$$x=\frac{2 \frac{1}{2\sqrt{g}} \sqrt{g}}{\sqrt{\frac{1}{2m}pCA}^2} \times ln|\frac{e^{\frac{\sqrt{\frac{1}{2m}pCA}}{ \frac{1}{2\sqrt{g}} }t}+1}{2e^{\frac{ \sqrt{\frac{1}{2m}pCA}}{2 \frac{1}{2\sqrt{g}} }t}}|$$
$$x=\frac{2m}{pCA} \times ln|\frac{e^{2 \sqrt{\frac{pCAg}{2m}}t}+1}{2e^{\sqrt{\frac{pCAg}{2m}}t}}|$$
____________________________________
Okay so can someone confirm the validity of my steps now, and confirm whether the below equation represents an object with mass m, dropped at a height of h, with respect to time t, given that gravity and drag are the only forces acting on it:

$$h=\frac{2m}{pCA} \times ln|\frac{e^{2 \sqrt{\frac{pCAg}{2m}}t}+1}{2e^{\sqrt{\frac{pCAg}{2m}}t}}|$$

 

[Theorγ]

I redid the problem on paper this time and I got a slightly different answer:

$$h=\frac{2mg}{pCA} \times ln|\frac{e^{2 \sqrt{\frac{pCAg}{2m}}t}+1}{2e^{\sqrt{\frac{pCAg}{ 2m}}t}}|$$

Compared to my original one from the above post:

$$h=\frac{2m}{pCA} \times ln|\frac{e^{2 \sqrt{\frac{pCAg}{2m}}t}+1}{2e^{\sqrt{\frac{pCAg}{ 2m}}t}}|$$

Can anyone confirm which one is right? My new one contains an extra g, but my other one doesn't...

 

[da_nang]

$$h=\frac{2m}{pCA} \times ln|\frac{e^{2 \sqrt{\frac{pCAg}{2m}}t}+1}{2e^{\sqrt{\frac{pCAg}{ 2m}}t}}|$$

It's this one. If you check the units, you get meter.