So first ball gets \(t \; sec\).

and \(2^{\text {nd }}\) gets \(( t -2) \; sec\) . and they will meet at same height

\(h _{1}=50 t -\frac{1}{2} gt ^{2}\)

\(h _{2}=50( t -2)-\frac{1}{2} g ( t -2)^{2}\)

\(h _{1}= h _{2}\)

\(50 t -\frac{1}{2} gt ^{2}=50( t -2)-\frac{1}{2} g ( t -2)^{2}\)

\(100=\frac{1}{2} g \left[ t ^{2}-( t -2)^{2}\right]\)

\(100=\frac{10}{2}[4 t -4]\)

\(5= t -1\)

\(t =6 \; sec\)