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(A) To solve this,we use the method of images. Reflect the billiard table across the two walls the ball hits. The path of the ball becomes a straight

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) To solve this,we use the method of images. Reflect the billiard table across the two walls the ball hits. The path of the ball becomes a straight line from $A$ to the image of pocket $B$ after two reflections.Let the horizontal distance from $A$ to the first wall be $(a-c)$. After the first reflection,the ball travels across the width $b$. After the second reflection,it travels a horizontal distance $a$. The total horizontal displacement required to reach the image of $B$ is $(a-c) + a = 2a - c$.The total vertical displacement is $2b$.From the geometry of the path,$	an 	heta = \frac{	ext{Total Vertical Displacement}}{	ext{Total Horizontal Displacement}} = \frac{2b}{2a - c}$.Therefore,$	heta = 	an^{-1} \left( \frac{2b}{2a - c} ight) = \cot^{-1} \left( \frac{2a - c}{2b} ight)$.

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

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