A body ‘A’ is projected upwards with a velocity of $\text{98}{\text{ms}}^{-1}$  and a second body B is projected upwards with the same initial velocity but after ‘4’ seconds. Both the bodies will meet after                                              

${\text{I}}_{\text{st}}$ Body$\text{u}=98{\text{ms}}^{-1}\text{at}\text{t}=0$  

${\text{II}}_{\text{nd}}$  Body $\text{u}=98{\text{ms}}^{-1}\text{at}\text{t}=4\text{s}$

Let the two bodies meat each other at t seconds  

First stone$\text{S}=98\text{t}-\frac{1}{2}{\text{gt}}^{\text{2}}\cdots \cdots \left(1\right)$

For second stone$\text{S}=98\left(\text{t}-\text{4}\right)-\frac{1}{2}\text{g}{\left(\text{t}-4\right)}^{\text{2}}\cdots \cdots \left(2\right)$

From (1) & (2)

$\Rightarrow 98\text{t}-\frac{1}{2}{\text{gt}}^{\text{2}}=98\text{t}-392\frac{1}{2}\text{g}\left({\text{t}}^{\text{2}}+16-2\left(\text{t}\right)\left(\text{u}\right)\right)$

$\Rightarrow 470.4=39.2\text{t}$

$\text{t}=12\text{s}$