Body 1: \(h=98 t-\frac{1}{2} \times 9.8 \times t^{2}\)

Body 2: \(h=98(t-4)-\frac{1}{2} \times 9.8 \times(t-4)^{2}\)

Equating the above two expressions we get:

\(98 t-\frac{1}{2} \times 9.8 \times t^{2}=98(t-4)-\frac{1}{2} \times 9.8 \times(t-4)^{2}\)

\(\Rightarrow 98 \times 4+\frac{9.8}{2}(16-8 t)=0\)

\(\Rightarrow 392=-4.9(16-8 t)\)

\(\Rightarrow \frac{392}{4.9}=(8 t-16) \Rightarrow 80=(8 t-16) \Rightarrow 96=8 t \Rightarrow t=12 s\)