[Physltest] [Phys-L] Re: projectiles

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding Jack's concession:

> Hi all-
>        Pls ignore my last posting, which was silly.  I have started
> off 2005 with a silly remark.  David is absolutely right.
>                                        Jack

I'm pleased that Jack has seen fit to see things my way on this.  But
I just now read his statement of acquiescence only *after* spending
some significant time on composing a careful explanation for the
correctness of my solution.  Since the explanation of the matter that
Jack brought up might be of interest to other Phys-l readers, and
since it is possible that they might not immediately see the problem
with Jack's objection, I beg Jack's indulgence (and that of the rest
of the Phys-l readership) and post my explanation below anyway.
-------------

Regarding:

> Hi David-
> I contend that the "solutions" cannot be equivalent because we each
> have an expression for Cos(2a).  These expressions must  be
> consistent with each other.  They are not.

They *are* consistent with each other.

>        The difficulty with your derivation is that it has
> introduced extraneous solutions.

No, it has not.  My derivation gives a unique value for s for each
legitimate value of q (i.e. any real value of q greater than -1).

> You can see this by noting that your definition of
> "s" gives it a range 0<s<2.  But s>1 in your first equation leads to
> equating a positive and negative number.
>                                        Regards,
>                                                Jack

No, it doesn't.  It equates a negative number to a negative number in
that case.  When s is between 1 and 2 (corresponding to a negative
value of cos(2*a) and a value of a between 45 and 90 degrees) the
value of q is *negative*.  It is this negative q-value that negates
my RHS when the LHS is negative.  IOW, the only time the optimal
launch angle is greater than 45 deg is when the launch position is
below the target height.  When q is positive and the launch is from
the top of a cliff rather than from the bottom of a valley a then the
optimum launch angle is less than 45 deg and both sides of the
equation are then positive.

BTW, Jack, your eqn. 2.) has the very same feature as mine does in
this regard (since mine is effectively a substitution into yours).

Recall your eqn 2.):

cos(2*a) = q*sin(a)/(sin(a) + sqrt(sin^2(a) + q))

*also* has the property that when cos(2*a) is negative the value of q
is also negative.  Recall that sin(a) is positive for all launch
angles that aim somewhat upward from horizontal.  Any range-extremal
solution that had a negative value of a and thus a negative value of
sin(a) would fire the projectile to the wrong zero crossing of the
parabola with the target height.  (It would effectively fire the
projectile backwards to the point that BC suggested to use as a
possible launch point that was on the parabola at the same height as
the target.)  This pathological case effectively changes the sign of
the sqrt() function in the denominator on the RHS and this
effectively switches which intersection the parabola makes with the
height of the target.  Once we dispose of the pathological negative
sin(a) case we see that your equation is equivalent to:

cos(2*a) = q/(1 + sqrt(1 + q/sin^2(a))).

Then once we make the substitution cos(2*a) --> 1 - s and
sin^2(a) --> s/2 we get the equation I started my derivation with,
i.e. :

1 - s = q/(1 + sqrt(1 + 2*q/s))   .

It is true that my derivation doesn't distinguish which algebraic
sign the sqrt() function has because of the step where I square both
sides of an equation that had the sqrt() isolated on one side.  But
this actually doesn't introduce any extraneous spurious solution
because the previously used restriction on the positivity of sin(a)
that canceled the original sin(a) factors in the RHS of the
starting equation.  After the equation-squaring step all the higher
order s-dependent terms that could potentially produce extra
solutions in s cancel out on both sides of the squared equation.  All
that survives this massive cancellation process is a *single*
s-dependent term that depends on just the (negative) first power of
s.  All the other surviving terms are s-independent.  This guarantees
that there can be no more than a single s-value (and thus a single
physical a-value) for each q-value.  Since there is only one s-value
for each q-value in my solution set, and since the original equation
(i.e. your eqn. 2.)) had a single iterative fixed-point solution for
a for each q-value we see that my derivation did not actually
introduce any extraneous spurious solutions.

David Bowman
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