Standing near the edge of the top of a building, a person throws...

Let the height of the building be $h$ and the initial speed of both balls be $u$.

For ball $A$, taking the downward direction as positive, the initial velocity is $-u$ and the displacement is $h$. Using the equation of motion:

$v_A^2 = (-u)^2 + 2gh = u^2 + 2gh$

For ball $B$, the initial velocity is $u$ downwards and the displacement is $h$. Using the equation of motion:

$v_B^2 = u^2 + 2gh$

From the above equations, $v_A^2 = v_B^2$, which implies $v_A = v_B$.

Alternatively, ball $A$ goes up and returns to the point of projection with the same downward speed $u$. From there, its motion is identical to that of ball $B$. Hence, both strike the ground with the same speed.