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...

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\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

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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:
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\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
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Using these variables as the following expressions:
\alpha = \frac{1}{2\sqrt{g}}
\beta = \sqrt{\frac{1}{2m}pCA}
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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
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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
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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}}|
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Previous Variable Denotation:
\alpha = \frac{1}{2\sqrt{g}}
\beta = \sqrt{\frac{1}{2m}pCA}
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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}}|
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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.