# Finding the equation for Velocity with Drag

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

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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.

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

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.

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:
$$\int\frac{a}{\sqrt{g}} ds= \int\tanh(\frac{bt}{2}) dt$$
$$\frac{a}{\sqrt{g}}s =\frac{2}{b} ln(cosh(\frac{bt}{2})) + Constant$$

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

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

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

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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:
$$\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} \times ln\left(\frac{|\sqrt{g}+v\sqrt{\frac{1}{2m}pCA}|}{ |\sqrt{g}-v\sqrt{\frac{1}{2m}pCA}|}\right)=t$$
$$\frac{\alpha}{\beta} \times ln\left(\frac{|\sqrt{g}+v\beta|}{ |\sqrt{g}-v\beta|}\right)=t$$
$$ln\left(\frac{|\sqrt{g}+v\beta|}{ |\sqrt{g}-v\beta|}\right)=\frac{\beta}{\alpha}t$$
$$e^{ln\left(\frac{|\sqrt{g}+v\beta|}{ |\sqrt{g}-v\beta|}\right)}=e^{\frac{\beta}{\alpha}t}$$
$$\frac{\sqrt{g}+v\beta}{\sqrt{g}-v\beta}=e^{\frac{\beta}{\alpha}t}$$

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

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Looks fine to me.

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:

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

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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...

$$h=\frac{2m}{pCA} \times ln|\frac{e^{2 \sqrt{\frac{pCAg}{2m}}t}+1}{2e^{\sqrt{\frac{pCAg}{ 2m}}t}}|$$
It's this one. If you check the units, you get meter.