How Does Air Resistance Affect the Velocity of a Falling Object?

[jfrank1034]

A ball of mass m is released from rest at x=0. Air resistance is expressed as R = kv² , where k is a positive constant and v denotes velocity.

Derive an expression for the speed in terms of the distance x that it has fallen.
Identify the terminal v.

I know I have to start with ma = mg - kv² and integrate but I can't seem to separate the variables right to get it as function of x or distance (even though the problem has a picture of a vertical fall)

 

[LawrenceC]

Hint: Express the 'a' in your equation differently.

 

[jfrank1034]

Okay so I have mv(dv/dx) = mg - kv² , and divide both sides by the (mg - kv²)

Then, (mv/(mg - kv²))dv = dx

∫(mv/mg - mv/kv²) dv = ∫dx

Just need a little help integrating if I'm on the right track I guess...

 

[LawrenceC]

Acceleration is not the derivative of velocity with respect to distance; it is the derivative with respect to time.

 

[jfrank1034]

Great...just realized that

 

[jfrank1034]

If I change a to d²s/dt² is this the right path?

 

[HallsofIvy]

Since your force is given in terms of v, it would be better to write the acceleration as dv/dt:
m\frac{dv}{dt}= mg- kv^2
and there is no problem with "separating variables"- you have
\frac{mdv}{mg- kv^2}= dt

 

[jfrank1034]

But I need the velocity in terms of the distance fallen, x. Integrating the right side, ∫dt , would just give me 't' am I right?

 

[jfrank1034]

Was I not correct in saying that a = v (dv/ds) ??

 

[jfrank1034]

In my book it says v dv = a ds --> a = v(dv/ds)...

so I have vdv/(-g + kv²) = dx

∫vdv/(-g + kv²) [from 0 to v] = ∫dx [from 0 to x]

(1/2k) ln(-g+kv²) [from 0 to v] = x

--> (1/2k) ln(g-kv²/g) = x

Solving for v and skipping a few steps I got v = √(g/k(1-e^2kx))

 

[LawrenceC]

Your units do not show velocity. You need a mass in there.

 

[jfrank1034]

You're right...should start as dx = mvdv/(-mg + kv²)

 

[LawrenceC]

Whenever you derive something, it is always prudent to check your units in the final product to see that they are what is intended. Secondly, it is a good idea to explore the limits of your formula to see if it makes sense at its limits.