Homework Help: Free fall with air drag force

1. Nov 6, 2004

UrbanXrisis

free fall with drag force

If the equation F=-bv^2 describes the drag force of an object...then the differential equation for the object's motion would be:

dv/dt= -g+bv/m

or is it...

dv/dt= g-bv/m

After solving the equation, should I get...
V=mg/b[1-e^(-bt/m)]

Also, does this look like the position v time graph of the object?

Last edited by a moderator: May 1, 2017
2. Nov 6, 2004

Pyrrhus

Ok i'll do the analysis.

Forces acting on a body falling.
Down positive (Using resistive force proportional to the speed)

We got a first order DE

$$mg - bv = m \frac{dv}{dt}$$

To not get into much detail, this type of DE

$$\frac{dy}{dt} = ay - b$$

Has the following solution

$$y = \frac{b}{a} + ce^{at}$$

For an initial value, to find C.

$$y = \frac{b}{a} + [y_{o} - \frac{b}{a}]e^{at}$$

Thus for our case the solution is

$$v = \frac{mg}{b} + [v_{o} - \frac{mg}{b}]e^{-\frac{bt}{m}}$$

If we arrange the terms and $v_{o} = 0$

$$v = \frac{mg}{b} - \frac{mg}{b}e^{-\frac{bt}{m}}$$

$$v = \frac{mg}{b}(1 - e^{-\frac{bt}{m}})$$

Last edited: Nov 6, 2004
3. Nov 6, 2004

UrbanXrisis

instead of F=-bv^2, what if F=-2v? The question tells me to ignore the effects of gravity in this problem. Would the DE then be dv/dt= 2v/m?

Since f=bv...and b=-2

dv/dt= g-bv/m
dv/dt= g-(-2)v/m
the question asks to ignore gravity...
dv/dt= -(-2)v/m
dv/dt= 2v/m

4. Nov 6, 2004

Pyrrhus

If we ignore gravity, then indeed our Newtonian analysis will be

The Object is falling, and the air drag is in the opposite direction
Down positive.

$$m\frac{dv}{dt} = -bv$$

$$\frac{dv}{dt} = \frac{-bv}{m}$$

Solving this:

$$\frac{dv}{dt} = \frac{-bv}{m}$$

$$\frac{dv}{v} = \frac{-b}{m}dt$$

$$\int_{v_{o}}^{v} \frac{dv}{v} = \int_{0}^{t} \frac{-b}{m}dt$$

$$\ln |v|]_{v_{o}}^{v} = \frac{-b}{m}t]_{0}^{t}$$

For $v_{o} = 0$

$$ln |v| = \frac{-b}{m}t$$

$$v = e^{\frac{-b}{m}t}$$

Last edited: Nov 6, 2004
5. Nov 6, 2004

UrbanXrisis

Does this look correct?

Last edited by a moderator: May 1, 2017
6. Nov 6, 2004

UrbanXrisis

However, the retarding force is not F=bv, it is F=-2v as stated above. Correct me if I’m wrong, but isn’t $$\frac{dv}{dt} = \frac{-bv}{m}$$ for when F=bv? Should it be $$\frac{dv}{dt} = \frac{2v}{m}$$ ?

7. Nov 6, 2004

Pyrrhus

Sure, just substitute b with 2, and for the graphic, It looks rather odd to me, but maybe someone else can find it the physical meaning.

Last edited: Nov 6, 2004
8. Nov 6, 2004

UrbanXrisis

my graph represents the object accelerating and to at point where it has constant velocity because of the drag force

or should it look something like this...

Last edited by a moderator: May 1, 2017
9. Nov 6, 2004

Pyrrhus

the graphic looks better now, and i added the extra solve steps for neglecting gravity.

10. Feb 10, 2010