So first ball gets $t \; sec$.

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

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

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

$h _{1}= h _{2}$

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

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

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

$5= t -1$

$t =6 \; sec$