The ball's velocity will be represented by the derivative of its distance function:
\(s'(t)=6t^2+6t\)
Now find the times \(t\) for which this is equal to 30, i.e. solve
\(6t^2+6t=30\implies t^2+t-5=0\)
This has two solutions, \(t=-\dfrac{1\pm\sqrt{21}}2\), but only one is positive and falls in the interval \([0,3]\). So the velocity reaches 30 cm/s when \(t=-\dfrac{1-\sqrt{21}}2\approx1.79\).
