Question

A billiard table whose length and width are as shown in the figure. A ball is placed at point $A$. At what angle ${}^{\prime }{q}^{\prime }$ 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).

• A. $\theta ={\cot }^{-1}\frac{2a-c}{2b}$
• B. $\theta ={\tan }^{-1}\frac{2a-c}{2b}$
• C. $\theta ={\cot }^{-1}\frac{c-a}{2b}$
• D. $\theta ={\cot }^{-1}\frac{c-a}{b}$

Answer 1

The correct option is A $\theta ={\cot }^{-1}\frac{2a-c}{2b}$

$y=a-b=a-(b-(a-c)\tan \theta )\cot \theta =a-b\cot \theta +a-c\tan \theta =\frac{b}{y}\tan \theta =\frac{b}{2a-c-b\cot \theta }(2a-c)\tan \theta -b=b\tan \theta =\frac{2b}{2a-c}\cot \theta =\frac{2a-c}{2b}\theta =\cot {}^{-1}\frac{2a-c}{2b}$