vertical motion under gravity, tricky question - help needed

Can anyone help my son find how to calculate the answer to this problem. We have been given the answer but can't arrive at it ...

A ball is thrown vertically upwards from a window 6m above the ground at a velocity of 8m per sec. 1.5 sec later a second ball is dropped from the same window. What distance below the window do they meet? I tried different methods, all gave different results, none of which was the answer given, 0.104 m. Can anyone help work this out please?

Can anyone help my son find how to calculate the answer to this problem. We have been given the answer but can't arrive at it ...

A ball is thrown vertically upwards from a window 6m above the ground at a velocity of 8m per sec. 1.5 sec later a second ball is dropped from the same window. What distance below the window do they meet? I tried different methods, all gave different results, none of which was the answer given, 0.104 m. Can anyone help work this out please?

Did you end up using two equations?

I used s=ut+1/2at(squared) twice, once for ball 1, and once for ball 2, letting t2 = t1-1.5, left me with two simultaneous equations containing s and t, solved them for t then worked out s. This may have been the wrong approach though.

This is the right approach, but I can imagine the signs getting in a tangle.

Since the balls will meet below the window, I suggest taking downwards as positive for both balls. This means that your s will turn out positive. For the first ball, u = - 8

This will give 0.1037.... for s