Vertical motion under gravity - help needed to find solution

The equation governing the motion for the balls is:
$s=ut+\frac{1}{2}at^2$
And bear in mind that the second ball is dropped 1.5s after the first, so $t_2=t_1-1.5$

We were unsure about your second point - although ball 2 is released 1.5 sec after ball 1, does it necessarily follow that its time in flight at the point the balls coincide is 1.5 sec less that that of ball 1 ...

Yes.

Initial velocities and acceleration are predetermined. The positions are therefore a function of time only. You describe the first ball's position, $s_1$ in terms of time $t_1$. You describe the second ball's position, $s_2$ in terms of time $t_2$. The balls meet when the positions are the same i.e. $s_1 = s_2$. This gives you an equation with the variables $t_1$ and $t_2$. How are these two related? Time progresses at a constant rate in any given inertial frame; the speed differences between the two balls is too insignificant to produce any noticeable relativistic effect on time. As such, the clock of $t_2$ starts ticking 1.5 seconds after that of $t_1$, and at the moment the balls meet, $t_2$ is still 1.5 seconds behind $t_1$