[Phys-L] Re: projectiles

[David_Bowman at GEORGETOWNCOLLEGE.EDU (David Bowman)]

Regarding Jack's contribution to the projectile discussion:

>                Here's a simple formula and a rapidly convergent
> algorithm for your case.  You can work it out easily for yourself if
> you just remember that horizontal distance traveled is a fine proxy
> for the elapsed time:
> 1       Define a parameter q=2gH/V^{2}  (V^{2} is the TeX way to
> write the exponent 2; V is the magnitude of the intitial velocity
> vector)
> 2.  The launch angle a for maximum range is given by:
>        Cos(2a)= qSin(a)/[Sin(a)+A)
> where A^{2} = Sin^{2}(a) +q
> 3.  For the algorithm, put a starting guess for a on the right hand
> side.
> The value you get for Cos(2a) gives you an improved value for a.  45
> degrees is a good starting guess.  This procedure converges to three
> places in one iteration for q=.4.  The angle for max range in that
> case is 40.2 degrees.

Jack, although the convergence rate is impressive, I don't understand
your reason for the iterative solution of your equation 2.) above.
The equation is explicitly solvable for the angle a. The solution is:

a = (1/2)*arcos(q/(2 + q))

If we use your parameter q = 0.4 = 2/5 the exact solution is
a = (1/2)*arcos(1/6) = 40.2029658866... deg.

Note that your parameter a is the same as my (& Anthony's) parameter
q.  Also, your parameter q is my parameter p and it physically is
the ratio of the initial potential energy (relative to the target
height) to the initial kinetic energy.

>        There may be a condition on the starting angle for values
> of q>3. The condition may be that Sin(a)<1/sqrt(q-2).  I'll leave it
> for others to explore this condition.

Any restriction on starting guesses for your iteration is an artifact
of the form and structure of the iteration equation itself.  There is
no restriction on positive q (my p) > 3 in that all positive q-values
give a well-defined unique acute solution value for a.  But there
*is* a physical restriction on the range of q (p) values when this
parameter is *negative*.  The condition is that we can only have a
solution if q > -1.  If q < -1 then the projectile does not have
enough energy to reach the height of the target (from below) and
there is no real range for any real launch angle that will get the
projectile that high for any part of its trajectory.

>                                        Regards,
>                                                Jack

David Bowman