પદાર્થ A ને ઉપર તરફ nbsp;98 m/s ના વેગથી

$h=u t-\frac{1}{2} g t^{2}$

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:

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

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

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

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