Re: Reach of a projectile

[John Mallinckrodt <ajm@CSUPOMONA.EDU>]

> Greetings folks!
>
> Consider the usual projectile motion problem in the vertical xy-plane=
> .
> A projectile is launched with speed v0 at an angle theta from the
> origin.
>
> If the variable theta is swept from 0 to 90 degrees what part of the
> xy-plane will be covered by the projectile?  What is the equation of =
> the
> boundry of this area?  I would also like to see a 3-D plot in which
> z-axis would represent the probability of the projectile passing thou=
> gh
> a point (x,y).  I believe it will peak at the origin.
>
> Is this worked out somewhere?  I am trying to solve it but the equati=
> ons
> are getting messier.  I am stil trying though.

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