a
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 \(v = \frac{{1 \times 10 \times 30 \times 6}}{{2 \times 12}} = 75\,m/s\)