[Phys-L] Re: projectiles

[David_Bowman at GEORGETOWNCOLLEGE.EDU (David Bowman)]

Regarding BC's seductive suggestion:

> this probably is off the wall, but could not one fire a projectile
> at 45 deg.  find what its angle is at height h and that be it?   For
> the speed it's the problem's given speed + what a rock would have at
> the ground dropped from h.
>
> bc, often crooked thinker

Nice try, but your suggestion is not correct in general.  This is
because maximizing the range of the remaining trajectory at an
arbitrary point in flight is not equivalent to maximizing the range
at a height that is equal to the target height.

If we define p == 2*g*h/v^2 to be the ratio of the initial potential
energy (above the target height) at launch to the initial kinetic
energy, then the maximal-range trajectory has an initial launch angle
from horizontal of (1/2)*arccos(p/(p + 2)) and has an impact angle
(relative to horizontal) at the target which is the *complement* of
this optimal launch angle.  Since the launch angle and the impact
angle are complements of each other for the maximal range trajectory,
and since the launch angle and the impact angle are equal (er,
congruent) to each other when the launch height is equal to the
target height (by the mirror symmetry of the parabolic trajectory),
this means that the common launch/impact angle is its own complement,
i.e. 45 deg, only for this special case.  If the launch height h (&
hence p) is something *other* than zero then the impact angle for the
maximal-range trajectory will *not* be 45 degrees, and thus the angle
of the back-extrapolated point on the maximal-range trajectory which
happens to be at the target's height will *also* *not* be 45 degrees
(by the mirror symmetry of the parabola) contrary to BC's suggestion.

Since the optimal launch and impact angles are mutual complements
this means that one of these angles is greater than 45 deg and the
other is less than 45 deg if they are not equal to each other.  It
ends up that when h (& thus p) is (are) positive the launch angle is
less than 45 deg and the impact angle is greater than 45 deg.  When
h (& p) is (are) negative then the situation is reversed.  This
negative h case can also be understood by using the positive h case
and then reversing the roles of the launch point and the impact point
and using the time-reversed trajectory.  The mapping for the reversal
is h --> -h and p --> -p/(1 + p).  Since the ratio of impact speed to
launch speed is sqrt(1 + p) this means the reversal mapping maps
1 + p --> 1/(1 + p).  *Only* when h = 0 (& p = 0) is this reversal
mapping an identity giving the launch and the impact the same angle
of 45 deg.

In terms of the (initial potential to kinetic energy) ratio p the
maximal range turns out to be R_max = sqrt(1 + p)*v^2/g and the range
R for generic launch angle q is
R = cos(q)*(sin(q) + sqrt(sin^2(q) + p))*v^2/g .

It is very useful to realize that (scaled versions of) both the
horizontal component of the momentum, i.e. v*cos(q) *and* the total
energy, i.e. (1 + p)*v^2 are conserved *throughout* the trajectory for
this problem.

David Bowman