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