The ball's position vector has components
[tex]x=\left(8.00\dfrac{\rm m}{\rm s}\right)\cos40.0^\circ t[/tex]
[tex]y=1.00\,\mathrm m+\left(8.00\dfrac{\rm m}{\rm s}\right)\sin40.0^\circ t-\dfrac g2t^2[/tex]
where [tex]g=9.80\dfrac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity. The ball hits the ground when [tex]y=0[/tex]:
[tex]0=1.00\,\mathrm m+\left(8.00\dfrac{\rm m}{\rm s}\right)\sin40.0^\circ t-\dfrac g2t^2\implies t=1.22\,\mathrm s[/tex]
The ball's velocity vector has components
[tex]v_x=\left(8.00\dfrac{\rm m}{\rm s}\right)\cos40.0^\circ[/tex]
[tex]v_y=\left(8.00\dfrac{\rm m}{\rm s}\right)\sin40.0^\circ-gt[/tex]
so that after 1.22 s, the velocity vector is
[tex]\vec v=(6.13\,\vec\imath-6.79\,\vec\jmath)\dfrac{\rm m}{\rm s}[/tex]
and the magnitude is
[tex]\|\vec v\|=\sqrt{6.13^2+(-6.79)^2}\,\dfrac{\rm m}{\rm s}=\boxed{9.14\dfrac{\rm m}{\rm s}}[/tex]
