[Phys-L] Re: thrown ball with air drag

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding the problem Carl mentioned:

> A ball is thrown vertically upward with initial speed v0. Assume air
> drag is proportional to speed squared. Let the speed of the ball
> just before it hits the ground be denoted v. Show that
> 1/v^2 = 1/v0^2 + 1/vT^2 where vT is the terminal speed of the ball.
>
> I feel like there must be some physical significance as to why this
> result comes out as a sum of inverse squares. (Doing the problem does
> not really shed any light on this interesting feature, at least the
> way I did it.) Anyone have any ideas about it?

This is indeed a remarkable result.  I have not yet thought of a
transparent reason (e.g. by some hidden symmetry or conservation
principle) for why that result should work out that way short of
actually solving the relevant DE and seeing that it just works out
that way.

That said, I have come up with a way to solve the DE in a relatively
low pain way that avoids a lot of the potential mess that attends a
naive solution.  The trimmed up method makes it particularly easy to
see how the result Carl alludes to comes about (at least compared
with a solution that involves a naive brute force solution of
Newton's 2nd law written in the original variables).  But getting
Carl's result *still* does seem to require that a DE be solved (even
though it happens to be fairly simple) for both the ascent and the
decent.

We start with the eqn. of motion:

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

where v is the upward (signed) velocity, v_t is the terminal speed
and s is the algebraic sign (+/-)1 depending on whether the object
is going up or going down.  If the object is ascending then s = +1,
and if the object is going down then s = -1.

By using the chain rule we can rewrite the acceleration in a way
that eliminates the unnecessary time parameter t, i.e.

dv/dt = v*dv/dy = (1/2)*d(v^2)/dy where v = dy/dt .  This makes the
eqn. of motion take the form:

d(v^2)/dy = -2*g*(1 + s*v^2/v_t^2) .

We next define two new dimensionless variables replacing the
dependent variable v and the independent variable y.  We let
r == v^2/v_t^2 be the object's kinetic energy in units of the
kinetic energy it would have if it moved at the terminal speed.
We also define z == 2*g*y/v_t^2 as a dimensionless height of the
object.  We can think of z as the object's potential energy in
units of the kinetic energy it would have if it moved at the
terminal speed.

Substituting these definitions into the eqn. of motion above results
in the new DE:

dr/dz = -(1 + s*r) =  -s*( r + s)  (because s^2 = 1) .  We can also

write:

d(r + s)/dz = -s*(r + s) or equivalently,

d ln(r + s)/dz = -s

whose solution trivially becomes:

ln((r+s)/(r_0 + s)) = -s(z - z_0) or equivalently,

r + s = (r_0 + s)*exp(-s*(z-z_0))                            eqn. 1

where r_0 and z_0 are the initial condition values of r & z
respectively.

Now we apply this solution separately to the two separate cases of
the object ascending and descending.  Suppose that the object is
initially launched with an upward velocity of v_i and that it
reaches a maximum height of H before beginning to fall back down.  We
also assume that the object at the moment of returning to the
original launching height has a final speed of v_f. We define
r_i == (v_i)^2/(v_t)^2, r_f == (v_f)^2/(v_t)^2 and h == 2*g*H/v_t^2.

Now consider the moment of reaching the top of the trajectory for the
upbound part of the motion.  In this case we have:

s = +1, r = 0, r_0 = r_i, z = h, and z_0 = 0.  Substituting these
particular values in to eqn. 1 gives:

1 = (r_i + 1)*exp(-h) or equivalently,

1 + r_i = exp(h) .                                           eqn. 2

Next we consider the moment the object returns to the launch height
on its downward path.  In this case we have that the parameters
become:

s = -1, r = r_f, r_0 = 0, z = 0, and z_0 = h.  Substituting these
particular values in to eqn. 1 gives:

r_f - 1 = -exp(-h) or equivalently,

1 - r_f = exp(-h) .                                           eqn. 3

We eliminate the common parameter h form both eqn. 2 and eqn. 3 by
multiplying these two equations together.  The result is:

(1 + r_i)*(1 - r_f) = 1 .

We next expand out the product and cancel the common unity term on
both sides of the equation, and then divide both sides by the product
r_i*r_f.  The result is

1/r_f - 1/r_i - 1 = 0 .

Rearranging this gives:

1/r_f = 1 + 1/r_i .                                          eqn. 4

Eqn. 4 is just Carl's beautiful result written in a dimensionless
form.

Since the result *is* so neat I suspect that the problem very well
may have some hidden symmetry or conservation law buried in it
somewhere.  Whatever it is (if it exists), it didn't jump put at me
yet.

David Bowman
_______________________________________________
Phys-L mailing list
Phys-L@electron.physics.buffalo.edu
https://www.physics.buffalo.edu/mailman/listinfo/phys-l