Let the two balls meet a fter t s at distance x from the platform.
x from the platforam
For the first ball
$u = 0,\,t = 18\,s,\,g = 10\,m/{s^2}$
Using  $h = ut + \frac{1}{2}g{t^2}$
$\therefore \,\,\,\,x = \frac{1}{2} \times 10 \times {18^2}$
For the second ball
$u = v,\,t = 12\,s,\,g = 10\,m/{s^2}$
Using $h = ut + \frac{1}{2}g{t^2}$
$\therefore x = v \times 12 + \frac{1}{2} \times 10 \times {12^2}$
From equations ($i$) and ($ii$), we get

$\frac{1}{2} \times 10 \times {18^2} = 12v + \frac{1}{2} \times 10 \times {\left( {12} \right)^2}$
or $12v = \frac{1}{2} \times 10 \times \left[ {{{\left( {18} \right)}^2} – {{\left( {12} \right)}^2}} \right]$
$ = \frac{1}{2} \times 10 \times \left[ {\left( {18 + 12} \right)\left( {18 – 12} \right)} \right]$
$12v = \frac{1}{2} \times 10 \times 30 \times 6$
or $v = \frac{{1 \times 10 \times 30 \times 6}}{{2 \times 12}} = 75\,m/s$