According to the second equation, of motion, in terms of initial velocity and uniform acceleration, $s=ut+\frac{1}{2}a{t}^2$, 
the displacement is given, 
$s=6 \mathrm{m}$, 

$$\begin{gathered} 6=4t+\frac{1}{2}4t^2 \\ \Rightarrow 2t^2+2t-6^2=0 \\ \Rightarrow t^2+2t-3=0 \end{gathered}$$

Now find the root of the above equation, 
$\Rightarrow t=\frac{-2\pm \sqrt{4+12}}{2}=1 \mathrm{s}$

then the formula, 
$v_x=4+4\times 1=8 \mathrm{m} {\mathrm{s}}^{-1}$, 
$v_y=0+4\times 1=4 \mathrm{m} {\mathrm{s}}^{-1}$

then the resultant velocity, 
$v=\sqrt{{\left(v_x\right)}^2+{\left(v_y\right)}^2}=\sqrt{8^2+4^2}=4\sqrt{5} \mathrm{m} {\mathrm{s}}^{-1}$