# Two Balls Colliding. Check my work please!

1. Sep 24, 2010

### mmmboh

[PLAIN]http://img178.imageshack.us/img178/4781/345b.jpg [Broken]

So I said v=v0-gt, and at the highest point, v=0, so t=v0/g.
I also said uy=u0sinx-gt, and ux=u0cosx.
So at t=v0/g, both balls have to be at their highest pint, and when uy=0, t=u0sinx/g...so equating the two times, I find u0sinx=v0...which I guess is obvious without calculation.
So uy0=v0, and ux0=v0cotx, and u0=v0/sinx.
In this time, the left ball must travel d, so ux0*t=v0cotx*v0/g=v02cotx/g=d...
I did some rearranging and found that v0=(d*g*tanx)1/2.
Since u0=v0/sinx=(d*g*tanx)1/2/sinx, we need to minimize (tanx)1/2/sinx= (1/(sinxcosx))1/2...which is a minimum at x=45o, and when x=45o, v=(d*g)1/2.
Is this right?

Last edited by a moderator: May 4, 2017
2. Sep 24, 2010

### kuruman

Looks right to me.