Two balls A and B of same masses are thrown from the top of the building. A, thrown upward with velocity V and B, thrown downward with velocity V, then

Let both the balls are thrown from height ‘h’ , both will move under same acceleration i.e., acceleration due to the gravity.

Let the speed of ball ‘A’ when it reach the ground is ${\mathrm{V}}_{\mathrm{A}}$

For ball ‘A’ using ${\mathrm{v}}^2= {\mathrm{u}}^2+2\mathrm{gh}$

${\left({\mathrm{V}}_{\mathrm{A}}\right)}^2 = {\mathrm{V}}^2+2\mathrm{gh} \Rightarrow {\mathrm{V}}_{\mathrm{A}} = \sqrt{{\mathrm{V}}^2+2\mathrm{gh}}$

Let the speed of ball ‘B’ when it reach the ground is ${\mathrm{V}}_{\mathrm{B}}$

For ball ‘B’  again using ${\mathrm{v}}^2 = {\mathrm{u}}^2+2\mathrm{gh}$

${\left({\mathrm{V}}_{\mathrm{B}}\right)}^2 = {(-\mathrm{V})}^2 +2\mathrm{gh} \Rightarrow {\mathrm{V}}_{\mathrm{B}} = \sqrt{{\mathrm{V}}^2+2\mathrm{gh}}$

So for both the cases velocity will be equal.