Answer
See below
Work Step by Step
If the rocket is launched from 5m above the ground, the only that changes from the previous question is the value of $c_2,c_2=5$
Obtain $y=\frac{1}{2}gt^2+c_1t+5$
then $90=\frac{1}{2}\frac{-c_1^2}{g}-\frac{c_1^2}{g}+5\\
90=-\frac{c_1^2}{2g}-\frac{c_1^2}{g}+5\\
85=-\frac{c_1^2}{2g}\\
c_1=\sqrt 170g\approx40.82 m/s$
For time, we have $t=-\frac{c_1}{g}=\frac{-40.82}{-9.8}=4.17$s
