With constant acceleration, we have
[tex]a=a_{\mathrm{av}}=\dfrac{v-v_0}t[/tex]
At the end of the throw, the ball's velocity is equal to its initial velocity in magnitude, but in the downward direction so its sign is opposite and [tex]v=-v_0[/tex]. Then
[tex]-9.85\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{-2v_0}{3.0\,\mathrm s}\implies v_0=15\,\dfrac{\mathrm m}{\mathrm s}[/tex]
