\(-h=-\frac{1}{2} g t^2\)  (\(1^{\text {st }} \text { ball }\))

\(\Rightarrow 122.5=\frac{1}{2} \times 9.8 t^2\)

\(\Rightarrow t^2=25 \Rightarrow t=5 \,s\)

Another ball is dropped after \(2\) second so it took only \((5-2)=3 s\)

\(-122.5=-u(3)-\frac{1}{2} \times 9.8 \times 3^2\)

\(\Rightarrow 122.5=3 u+4.9 \times 9\)

\(\Rightarrow 3 u=78.4\)

\(\Rightarrow u=26.1 \,s\)