Let's play Football

First of all , due to impulse

Horizontal velocity $Vx = 6 m/sec$

Vertical velocity $Vy = 8 m/sec$

Angular velocity $\omega = 125 rad/sec$

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Before the first collision ,range travelled is $R1 = 9.6m (<10m)$

Since it is less than 10m , hence it is still on the smooth surface, friction is 0. (Normal collision)

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In between first and the second collision , it travelled a further range of $R2 = 4.8 m$

Now the friction is there.... It is sufficient to cause pure rolling

$F dt = m ( v - Vx )$

$FR dt = \frac{2mR^2 (125 - \omega )}{5}$

And ... $\omega R = v$

You will get $v = \frac{80}{7}$

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In between the second and the third collision, range travelled is $R3 = \frac{32}{7}$

Now , since their is no slipping therefore no frictional impulse will act for the successive collision(i.e. For the third , fourth and so on)

Range forms the series R3 + R4 + R5 + .... = $\frac{32}{7} ( 1 + e + e^2 + e^3 + .....) = 9.142$

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Hence final displacement is $R1 + R2 + R3 + .... = 9.6 + 4.8 + 9.142 = \boxed{23.542}$