[tex]\bf s(t)=-16t^2+144\implies \stackrel{velocity}{\cfrac{ds}{dt}}=-32t[/tex]
when does it hit the ground? check the picture below.
[tex]\bf 0=-16t^2+144\implies 16t^2=144\implies t^2=\cfrac{144}{16}\implies t^2=9
\\\\\\
t=\sqrt{9}\implies t=3[/tex]
at that moment, the velocity is -32(3).