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(C) For a wheel rolling without slipping with speed $v$,the velocity of the highest point $B$ is $2v$ in the horizontal direction.When the mud separat

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$4v\sqrt{\frac{r}{g}}$

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(C) For a wheel rolling without slipping with speed $v$,the velocity of the highest point $B$ is $2v$ in the horizontal direction.When the mud separates at point $B$,it acts as a projectile with an initial horizontal velocity $u_x = 2v$ and initial vertical velocity $u_y = 0$.The height of point $B$ from the ground is $2r$.Using the equation of motion for vertical displacement: $h = u_y t + \frac{1}{2} g t^2$,we have $2r = 0 + \frac{1}{2} g t^2$.Solving for time $t$,we get $t = \sqrt{\frac{4r}{g}} = 2\sqrt{\frac{r}{g}}$.The horizontal distance covered by the mud during this time is $AC = u_x 	imes t$.Substituting the values,$AC = (2v) 	imes (2\sqrt{\frac{r}{g}}) = 4v\sqrt{\frac{r}{g}}$.

$A$ wheel of radius $r$ rolls without slipping with a speed $v$ on a horizontal road. When it is at a point $A$ on the road,a small piece of mud separates from the wheel at its highest point $B$ and drops at point $C$ on the road. The distance $AC$ will be

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