Answer
$3.53~m/s^{2}$, north.
Work Step by Step
Let north be the positive direction:
$v_{0}=-81.9~m/s$ and $v_{f}=0$
$v_{av}=\frac{1}{2}(v_{f}+v_{i})=\frac{1}{2}[0+(-81.9~m/s)]=-40.95~m/s$
$x=x_{0}+(v_{av})t$. But, $x_{0}=0$ and $x=-949~m$ (origin in the landing position):
$-949~m=(-40.95~m/s)t$
$t=\frac{-949~m}{-40.95~m/s}=23.17~s$.
Finally:
$a_{av}=a=\frac{\Delta v}{\Delta t}=\frac{v_{f}-v_{i}}{t}=\frac{0-(-81.9~m/s)}{23.17~s}=+3.53~m/s^{2}$
+ means north.
