Two balls A and B are placed at the top of 180   m tall tower. Ball A is released from the top at t = 0…

Displacement covered by first ball,$\left(180-100\right)=0+\frac{1}{2}\times 10\times t^2$   $\Rightarrow t=4 \mathrm{s}$

Now, the second body gets only,  $t_2=4-2=2 \mathrm{s}$.

Displacement covered by second will be same,

$80=u\times 2+\frac{1}{2}\times 10\times {\left(2\right)}^2$

$\Rightarrow u=\frac{80-20}{2}=30 \mathrm{m} {\mathrm{s}}^{-1}$