Answer
$v=\sqrt (v^2_x+v^2_y)=\sqrt (v^2_{x_0}+v^2_{y_0})=v_{0}$
Work Step by Step
During a course of motion, the horizontal component of the speed does not change, so $v_x=v_{x_0}$. If the firing level equals the landing level, the net vertical displacement is 0, so $y-y_0=0$. Thus,
$$v^2_y=v^2_{y_0}+2a_y(y-y_0)=v^2_{y_0}$$
$$v^2_y=v^2_{y_0}, v^2_x=v^2_{x_0}$$
The initial speed is $v_0=\sqrt (v^2_{x_0}+v^2_{y_0})$
The final speed is $v=\sqrt (v^2_x+v^2_y)=\sqrt (v^2_{x_0}+v^2_{y_0})=v_{0}$
Thus, $v=v_0$.
