Bouncing Between Rings

The collision is elastic with no other forms of resistance , so the ball will bounce with the same magnitude of horizontal velocity $v_x$
the magnitude of the change in the horizontal component of the ball’s momentum , $\Delta p=2mv_x$

$d=2r(1-cos\theta)$

$t=\frac{x(t)}{v_x}$ $\text{........................(1)}$

$y(t)=v_yt-\frac{1}{2}gt^{2}$ $\text{..........(2)}$

at $t=\frac{d}{v_x}$ , $y(t) =0$ , ... sub. into eq.(2) ,

$0=v_y\frac{d}{v_x}-2g(\frac{d}{v_x})^{2}$

$v_x=\frac{gd}{v_y}=\frac{2gr(1-cos\theta)}{2vsin\theta} =\frac{gr(1-cos\theta)}{vsin\theta}$

but $v=\frac{v_x}{cos\theta}$

$(v_xcos\theta)^{2}=\frac{gr(1-cos\theta)}{tan\theta}$

Now, to find the magnitude of angle $\theta$ at which $v_x$ is maximum we differentiate the above equation W.R.T $\theta$ and equate the result to zero ,

$\frac{d(v_xcos\theta)^{2}}{d\theta} =gr\frac{ sin\theta\ tan\theta - (1-cos\theta) sec^{2}\theta}{tan^{2}\theta} = 0$

$sin\theta\ tan\theta - (1-cos\theta) sec^{2}\theta = 0$

$(1-cos\theta)(cos\theta(1+cos\theta)-1)=0$

$\text{Either}$ $(1-cos\theta)=0$ , $\theta=0 , \pi , 2\pi , ...$ $\text{Neglected}$

$\text{Or}$ $cos\theta(1+cos\theta)-1=0$ ==> $\theta = \frac{\sqrt{5}-1}{2} \approx\ 0.618$ $\text{YES , we can see the finger print of Golden ratio}$