Question
A billiard table whose length and width are as shown in the figure. $A$ ball is placed at point $A$. At what angle ‘$\theta $ ’the ball be projected so that after colliding with two walls, the ball will fall in the pocket $B$ .Assume that all collisions are perfectly elastic (neglect friction)

Answer

$y=a-b$

$=a-(b-(a-c) \tan \theta) \cot \theta$

$=a-b \cot \theta+a-c$

$\tan \theta=\frac{b}{y}$

$\tan \theta=\frac{b}{2 a-c-b \cot \theta}$

$(2 a-c) \tan \theta-b=b$

$\tan \theta=\frac{2 b}{2 a-c}$

$\cot \theta=\frac{2 a-c}{2 b}$

$\theta=\cot ^{-1} \frac{2 a-c}{2 b}$

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