The ball rolls off the table with speed <em>v</em> from a height of 0.35 m, so that it covers a horizontal distance <em>x</em> with height <em>y</em> at time <em>t</em> of
<em>x</em> = <em>v t</em>
<em>y</em> = 0.35 m - 1/2 <em>g t</em> ²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.
Solve for <em>t</em> when <em>y</em> = 0, i.e. the time it takes for the ball to reach the ground:
0 = 0.35 m - 1/2 <em>g</em> <em>t</em> ²
<em>t</em> ² = (0.70 m) / <em>g</em>
<em>t</em> ≈ 0.267 s
Now solve for <em>v</em> given that the ball falls 3 m away from the table:
3 m = <em>v</em> (0.27 s)
<em>v</em> = (3 m) / (0.27 s)
<em>v</em> ≈ 11.2 m/s