Given:
As long as angle beta is small, we have a right angle at the base of the bad guys flagpole.
Bad\;guys\;flagpole = 30\;metersAngle\;that\;range\;is\;optimized\; = 45\;degrees
Bad\;guys\;cannon\;lobs\;cannon\;ball\;at\;velocity\; = 56 \frac{meters}{second}
\sin{45 {degrees}} = 0.707106781
g = 10.0 \frac{meters}{seconds^2}
V_{y-at-peak} = 0
Reasoning
At\;45\;degrees,\;both\;vertical\; and\;horizontal\;velocities = 56 * \sin({45 {degrees}})
Solving:
V_{0-X-cannon} = V_{0-Y-cannon} = 56\;\frac{m}{s} * 0.707106781 = 39.597979736
t_{up-cannon} = \frac{(V_{y-at-peak} - V_{0-Y-cannon})}{g}
t_{up-cannon} = \frac{(0 - 39.597979736)}{10} = 3.597979735999997\;seconds
t_{total-up-down} = 2 * t_{up-cannon} = 7.9195959471999995
t_{total} = t_{total-up-down} = t_{horizontal} = 7.9195959471999995
x_{distance} = V_{0-X-cannon} * t_{total}
x_{distance} = 39.597979736 * 7.9195959471999995 = 313.5999998345333\;meters
Now we have a triangle: %vertical = 30\; m\; and\;horizonal = 313.5999998345333\;m angle_\beta = arctan(30/313.5999998345333) angle_\beta = 5.46447254\;degrees$
5.46447254 degrees is as close as our boat can safely go.
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