Re: Reach of a projectile

[Brian Whatcott <betwys1@SBCGLOBAL.NET>]

Hasan,

your proposal makes sense to me. It sounded as though the original
response was intended to maximize Y with theta - in fact I placed
that text in bold.

Brian W

At 07:34 AM 12/17/2003, you wrote:
> John,
> I am able to derive the equation y = A - Bx^2 that you mentioned.
> However, I still have a question:  How can we be sure that maximizing
>  y  wrt theta will span all the points covered by the projectile?
> What if I maximize r wrt theta where (r,theta) is
> the polar coordinate of a trajectory?
>
> Hasan Fakhruddin
> I've seen this written up somewhere in the past almost certainly



> TPT   I would imagine.
>
> First you find
>
>    y =  x*tan(theta) - g*x^2/(2*v0^2*(cos(theta))^2)
>
> then maximize y wrt theta and plug the answer back into this equation
> .
> The result is the upper envelope of all points that can be reached by
> the projectile and the answer is the parabola
>
>    y = A - B*x^2
>
> where A = v0^2/(2*g) and B = g/(2*v0^2).
>
> --
> John Mallinckrodt        mailto:ajm@csupomona.edu
> Cal Poly Pomona          http://www.csupomona.edu/~ajm


Brian Whatcott    Altus OK    Eureka!