Re: Reach of a projectile

[Roger Haar <haar@PHYSICS.ARIZONA.EDU>]

Hasan,
        It seems to me that for a given x maximizing y is
the same as maximmizing r = sqrt( x^2 + y^2)?

Tnaks,
Roger Haar

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"Fakhruddin, Hasanbhai" wrote:
> John,
> I am able to derive the equation y = A - Bx2 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
> Instructor of Physics
> The Indiana Academy for Science, Mathematics, and Humanities
> BSU
> Muncie, IN 47306
> E-mail: hfakhrud@bsu.edu
>
> 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