Question

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 $8m$ apart with their highest points at height $32m$ 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

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

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