A small ball is to be thrown, so as to just pass through three equal rings of diameter 2m and placed in parallel vertical planes at distance 8 m apart with their highest points at height 32 m above the point of projection as shown in the figure. If the angle of projection with horizontal is $\alpha$. The value of tan $\alpha$ is ______.

In figure, points A,B and C are lying on the trajectory of the path of the ball. The equation of trajectory is

$y=x\tan \alpha -\frac{g{x}^2}{2u^2{\cos }^2\alpha }$

For point A,

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}32-2=x\tan \alpha -\frac{5x^2}{u^2{\cos }^2\alpha }\\ \text{ Or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}30=x\tan \alpha -\frac{5}{u^2}{x}^2\left(1+{\tan }^2\alpha \right) ......\left(\mathrm{i}\right)\end{array}$

For point B,

$32=(x+8)\tan \alpha -\frac{5}{u^2}(x+8{)}^2\left(1+{\tan }^2\alpha \right) .......(\mathrm{ii})$

For point C,

$\begin{array}{l}32-2=(x+8)\tan \alpha -\frac{5(x+16{)}^2}{u^2}\left(1+{\tan }^2\alpha \right)\\ \Rightarrow 30=(x+16)\tan \alpha -\frac{5(x+16{)}^2\left(1+{\tan }^2\alpha \right)}{u^2} .....\left(\mathrm{iii}\right)\end{array}$

Subtracting Eq. (i) from Eq. (ii),

$\begin{array}{l}2=8\tan \alpha -\frac{5}{u^2}(1+\tan \alpha )\left(x^2+16x+64-x^2\right)\\ \Rightarrow 2=8\tan \alpha -\frac{5}{u^2}\left(1+{\tan }^2\alpha \right)(16x+64) .......\left(iv\right)\end{array}$

Subtracting Eq. (i) from Eq (iii),

$0=16\tan \alpha -\frac{5\left(1+{\tan }^2\alpha \right)}{u^2}\left(x^2+32x+256-x^2\right) ....\left(v\right)$

From Eq. (iv) and Eq.   

$\begin{array}{l}{\tan }^2\alpha =\frac{32\times 2}{64}=1\\ \therefore \tan \alpha =1\end{array}$