Unit 2, Problem 6 - Safety on a ship

Given:

As long as angle beta is small, we have a right angle at the base of the bad guys flagpole.

$$Bad\;guys\;flagpole = 30\;meters$$
$$Angle\;that\;range\;is\;optimized\; = 45\;degrees$$
$$Bad\;guys\;cannon\;lobs\;cannon\;ball\;at\;velocity\; = 56 \frac{meters}{second}$$
$$\sin{45 {degrees}} = 0.707106781$$
$$g = 10.0 \frac{meters}{seconds^2}$$
$$V_{y-at-peak} = 0$$

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.