- #1

mmmboh

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

So I said v=v

I also said u

So at t=v

So u

In this time, the left ball must travel d, so u

I did some rearranging and found that v

Since u

Is this right?

So I said v=v

_{0}-gt, and at the highest point, v=0, so t=v_{0}/g.I also said u

_{y}=u_{0}sinx-gt, and u_{x}=u_{0}cosx.So at t=v

_{0}/g, both balls have to be at their highest pint, and when u_{y}=0, t=u_{0}sinx/g...so equating the two times, I find u_{0}sinx=v_{0}...which I guess is obvious without calculation.So u

_{y0}=v_{0}, and u_{x0}=v_{0}cotx, and u_{0}=v_{0}/sinx.In this time, the left ball must travel d, so u

_{x0}*t=v_{0}cotx*v_{0}/g=v_{0}^{2}cotx/g=d...I did some rearranging and found that v

_{0}=(d*g*tanx)^{1/2}.Since u

_{0}=v_{0}/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=45^{o}, and when x=45^{o}, v=(d*g)^{1/2}.Is this right?

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