The ball traverses distance upwards and then downwards.

Hence, time taken is going from $\text{A}$ to $\text{B}$ is $\frac{\text{t}}{2}=\frac{4}{2}=2\text{s}$

Also at the maximum height $\text{B}$, speed is zero, therefore

$v=u-g t_{\max }$

where $v=0, t_{\max }=2s$, $g$ $\text{= 10}{\text{m/s}}^2$

$\therefore$ $u=g t_{\max }$ $\text{= 10}\times 2=20\text{m/s}$