Given:-

For a vertical motion ( $\uparrow$ ve)

$t=6s, a=- g=-9.81m/s^2$

To find:-

(i) Maximum Height $(h=y_{max})$

(ii)Initial velocity (u)

Solution:

For motion from point 1-2

$t=3s\space\space\space (\text {By symmetry }) \\ v=0 \\ a=-9.81m/s^2$

$u=?$

Using $v=u+at \\ \therefore 0=u-9.81×3 \\ \therefore u=29.43m/s \Rightarrow Ans \\ \text {Using } v^2=u^2+2as \\ \therefore 0=(29.43)^2-2×9.81×h \\ \therefore h= \dfrac {(29.43)^2}{2×9.81} \\ \therefore h=44.145m \Rightarrow Ans$