A carrom board ( 4 ft × 4 ft square) features the queen at its...

Let the carrom board be placed in the $xy$-plane with its corners at $(0,0)$, $(4,0)$, $(4,4)$, and $(0,4)$.

The centre of the board where the queen is initially placed is at $(2,2)$.

Let the front edge be the line $y=4 and the hole behind the striking line be at the origin$(0,0)$.<br><br>Let the queen strike the front edge at a point$(x,4)$.<br><br>According to the law of reflection, the angle of incidence equals the angle of reflection. This implies that the ratio of the horizontal distances covered is equal to the ratio of the vertical distances covered.<br><br>Vertical distance from the centre to the front edge is$4 - 2 = 2 \text{ ft}$.<br><br>Vertical distance from the front edge to the hole is$4 - 0 = 4 \text{ ft}$.<br><br>Horizontal distance from the centre to the point of impact is$2 - x$.<br><br>Horizontal distance from the point of impact to the hole is$x - 0 = x$.<br><br>Equating the ratios of horizontal to vertical distances:<br><br>$\dfrac{2 - x}{2} = \dfrac{x}{4}$<br><br>$4 - 2x = x \Rightarrow 3x = 4 \Rightarrow x = \dfrac{4}{3} \text{ ft}$<br><br>The point of impact on the front edge is \left(\dfrac{4}{3}, 4\right)$.

The displacement of the queen from the centre to the front edge is:

$d = \sqrt{\left(\dfrac{4}{3} - 2\right)^2 + (4 - 2)^2}$

$d = \sqrt{\left(-\dfrac{2}{3}\right)^2 + 2^2}$

$d = \sqrt{\dfrac{4}{9} + 4} = \sqrt{\dfrac{40}{9}} = \dfrac{2\sqrt{10}}{3} \text{ ft}$