## Essential University Physics: Volume 1 (4th Edition)

$83.5\ m$
We use the equation $x=v_{x0}t+\frac{1}{2}a_xt^2$ to get that the change in x is 79 meters. We use the equation $y=v_{y0}t+\frac{1}{2}a_yt^2$ to find that the change in y is 27 meters. Thus, we find that the value of r, the displacement, is: $r = \sqrt{79^2+27^2}\approx 83.5\ m$