For the first ball falling from the top of building, $u_1=0, u_2=40 \mathrm{~m} / \mathrm{s}, h=45 \mathrm{~m}, a_1=g, t_1=t, v=v_1$ Velocity of the dropped ball after time $t$,
$$
\begin{aligned}
&v_1=u_1+a_1 t \\
&v_1=u_1+g t \\
&\Rightarrow v_1=g t
\end{aligned}
$$
(downward)
For the second ball thrown upward,
$$
u_2=40 \mathrm{~m} / \mathrm{s}, a_2=-g, v=v_2, t_2=t
$$
Velocity of the ball after time $t$,
$$
\begin{aligned}
&v_2=u_2+a_2 t \\
&v_2=u_2-g t \\
&v_2=(40-g t) \\
&\text { (upward) } \\
&v_2=-(40-g t) \quad \text { (downward) } \\
&\therefore \text { Relative velocity of Ist ball w.r.t. IInd ball } \\
&v_2=v_1-v_2 \\
&=g t-[-(40-g t)]=40 \mathrm{~m} / \mathrm{s} \quad \text { (downward) } \\
&
\end{aligned}
$$
So their relative speed remain $=(40-0)=40 \mathrm{~m} / \mathrm{s}$.