We know that, coefficient of restitution,
$e=\left(\frac{\text { Relative velocity of separation }}{\text { Relative velocity of approach }}\right)$
Now, given collision is as shown


As impact is along normal to surface it does not change the horizontal component of ball.
Now if $v_{2 y}$ is final vertical component then
$v_{2 y}=e . v_{1 y}=0.2 \times 5 \cos 60^{\circ}$
or $\quad v_{2 y}=0.2 \times 5 \times \frac{1}{2}=0.5 \mathrm{~m} / \mathrm{s}$
So components of final velocity of ball are
$v_{2 x}=v_{1 x}=5 \sin 60^{\circ}=(5 \sqrt{3}) / 2$
and $v_{2 y}=e \cdot v_{1 y}=1 / 2$
Hence, final speed of ball, $v_f=\sqrt{\left(v_{2 x}\right)^2+\left(v_{1 x}\right)^2}$
$\begin{aligned} & =\sqrt{\left(\frac{5 \sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2} \\ & =\sqrt{\frac{75}{4}+\frac{1}{4}}=\sqrt{\frac{76}{4}}=\sqrt{19} \mathrm{~m} / \mathrm{s} .\end{aligned}$