Answer
$t\approx5.44s$
Work Step by Step
The path of the package can be divided into two parts: when the package is traveling upwards and when the package is traveling downwards.
The maximum height of the package can be calculated with the equation $v^{2}_{f}=v^{2}_{0}+2ax$. The velocity at the maximum height is $0m/s$ so the final velocity term reduces to $0$. Then solving for $x$, we get $x=\frac{-v_{0^{2}}}{2g}=\frac{-(12m/s)^{2}}{2(-9.8m/s^{2})}\approx7.35m$ (This is the maximum height above $80m$, so the actual maximum height is $87.35m$).
The time taken for the package to reach its maximum height can be calculated with the formula $d=\frac{1}{2}at^{2}$. Solving for $t$, we get $t=\sqrt {\frac{2x}{g}}$. The calculation for the time to reach the maximum height is $t_{1}=\sqrt {\frac{2(7.35m)}{9.8m/s^{2}}}\approx1.22s$. The time for the package to reach the ground from the maximum height can be calcualted using the same formula: $t_{2}=\sqrt {\frac{2(87.35m)}{9.8m/s^{2}}}\approx4.22s$.
The total time is the sum of $t_{1}$ and $t_{2}$:
$t=(1.22s)+(4.22s)=5.44s$
