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*From*: David Bowman <David_Bowman@georgetowncollege.edu>*Date*: Sun, 13 Jan 2019 17:56:50 +0000

Brian, the DE in question is fully separable and integrable with a fairly simple analytic solution (i.e. a hyperbolic tangent function). Once we redefine the problem in terms of dimensionless variables the solution is even universal.

Let a downward motion be the positive direction. The DE can be written as

m*dv/dt = m*g*(1 - (v/v_t^2))

where the terminal speed v_t is given by v_t = sqrt(2*m*g/(C_d*[rho]*A)). If we define a dimensionless dependent variable as u ==v/v_t, and define [tau] == g*t/v_t as a dimensionless independent temporal variable then the DE in terms of u & [tau] becomes: du/d{tau] = 1 - u^2 . This separable equation integrates to

u = tanh([tau] - [tau]_0) where [tau]_0 is an initial condition dependent integration constant. Translating this solution back to the original variables v & t gives

v = v_t*tanh(C*t + D) where

C == sqrt(C_d*[rho]*A*g/(2*m)) = g/v_t and

D == arctanh(v_0/v_t) .

Here the initial falling velocity v_0 is assumed to be *positive*, so the object is already headed downward somewhat initially when t = 0. OTOH if v_0 is negative and the particle is initially launched upward then things are significantly more complicated because for the upward portion of the motion the drag force reverses direction and this changes the DE so the solution for the upbound segment is written in terms of trig functions instead of hyperbolic functions. The overall solution then has a splice in it matching the upward motion to the downward motion at the moment when v = 0 at the peak of the ascent.

BTW, if the object is launched straight upward from some given level at a speed v_u, and it is moving at speed v_d on the way down at the moment when it reaches the same level, then v_u, v_d and v_t are mutually related by

1/(v_d)^2 = 1/(v_u)^2 + 1/(v_t)^2 .

David Bowman

________________________________________

From: Phys-l <phys-l-bounces@mail.phys-l.org> on behalf of brian whatcott <betwys1@sbcglobal.net>

Sent: Saturday, January 12, 2019 10:21 PM

To: prefered phys-l address

Subject: [Phys-L] Falling - the tale of a simple 1st degree non-linear ODE

I am pleased to use some of my dotage in answering questions on Quora.

I found this one strangely interesting:

How long does it take a jumper to reach terminal velocity?

An expression for the speed at which air drag equals weight goes like this:

jumpers mass times acceleration equals the net force provided by weight,

namely jumper's mass times g less air drag, namely 1/2 rho Cd A v^2

I spent way too much time in basing an estimate on the CR charging curve.

Apart from the similar curve, the rise time features differ greatly.

Casting around for better methods to plot this curve ~

Plugging the equation into Alpha solver on line?

A few lines of code in basic?

Mathematica?

Matlab?

It may be my imagination but the free use of Alpha online seems to be

more constrained than it used to be.

I could extract a plot, but the plotting was hit or miss - depending on

the coefficients I chose.

Basic code could provide a list of v against t but a good plot was lacking.

Mathematica was uncooperative, perhaps because I rarely use it.

Matlab had the ODE solvers but the examples were unhelpful.

Finally I looked up an MIT tutorial on using MATLAB with an example ODE

recipe which worked with my package, and took it from there.

https://i.imgur.com/K7KKlUG.jpg Matlab ODE

https://imgur.com/Eio6xs3 Alpha online

Brian W

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**Follow-Ups**:**Re: [Phys-L] Falling - the tale of a simple 1st degree non-linear ODE***From:*Richard Tarara <rtarara@saintmarys.edu>

**References**:**[Phys-L] Falling - the tale of a simple 1st degree non-linear ODE***From:*brian whatcott <betwys1@sbcglobal.net>

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