Two balls A and B are placed at the top of 180m tall tower. Ball A...

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

Now, the second body gets only, ${t}_{2} = 4 - 2 = 2s$.

Displacement covered by second will be same,

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

$⇒ u = \dfrac{80 - 20}{2} = 30m{s}^{- 1}$