Question

A small ball is to be thrown, so as to just pass through three equal rings of diameter $2 \,m$ 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 _____.

Answer 1

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

$y=x \tan \alpha-\frac{g x^{2}}{2 u^{2} \cos ^{2} \alpha}$
For point $A$,
$32-2=x \tan \alpha-\frac{5 x^{2}}{u^{2} \cos ^{2} \alpha}$
Or $30=x \tan \alpha-\frac{5}{u^{2}} x^{2}\left(1+\tan ^{2} \alpha\right)$ ....(i)
For point $B$,
$32=(x+8) \tan \alpha-\frac{5}{u^{2}}(x+8)^{2}\left(1+\tan ^{2} \alpha\right)$ ....(ii)
For point $C$,
$32-2=(x+8) \tan \alpha-\frac{5(x+16)^{2}}{u^{2}}\left(1+\tan ^{2} \alpha\right) $
$\Rightarrow 30=(x+16) \tan \alpha-\frac{5(x+16)^{2}\left(1+\tan ^{2} \alpha\right)}{u^{2}}$ .....(iii)
Subtracting Eq. (i) from Eq. (ii),
$2 =8 \tan \alpha-\frac{5}{u^{2}}(1+\tan \alpha)\left(x^{2}+16 x+64-x^{2}\right) $
$\Rightarrow 2 =8 \tan \alpha-\frac{5}{u^{2}}\left(1+\tan ^{2} \alpha\right)(16 x+64)$....(iv)
Subtracting Eq. (i) from Eq (iii),
$0=16 \tan \alpha-\frac{5\left(1+\tan ^{2} \alpha\right)}{u^{2}}\left(x^{2}+32 x+256-x^{2}\right)$ ....(v)
From Eq. (iv) and Eq. (v)
$ \tan ^{2} \alpha=\frac{32 \times 2}{64}=1 $
$\therefore \tan \alpha=1$