Answer
As $v$ increases from $43~ft/s$ to $45~ft/s$, the distance increases from $64.4 ~ft$ to $69.9 ~ft$. As $v$ increases, the distance increases significantly.
Work Step by Step
$D = \frac{v^2~sin~\theta~cos~\theta+v~cos~\theta~\sqrt{(v~sin~\theta)^2+64~h}}{32}$
We can find $D$ when $v = 43~ft/s$:
$D = \frac{(43~ft/s)^2~sin~42^{\circ}~cos~42^{\circ}+43~ft/s~cos~42^{\circ}~\sqrt{(43~ft/s~sin~42^{\circ})^2+(64)(7~ft)}}{32}$
$D = 64.4~ft$
We can find $D$ when $v = 44~ft/s$:
$D = \frac{(44~ft/s)^2~sin~42^{\circ}~cos~42^{\circ}+44~ft/s~cos~42^{\circ}~\sqrt{(44~ft/s~sin~42^{\circ})^2+(64)(7~ft)}}{32}$
$D = 67.1~ft$
We can find $D$ when $v = 45~ft/s$:
$D = \frac{(45~ft/s)^2~sin~42^{\circ}~cos~42^{\circ}+45~ft/s~cos~42^{\circ}~\sqrt{(45~ft/s~sin~42^{\circ})^2+(64)(7~ft)}}{32}$
$D = 69.9~ft$
As $v$ increases from $43~ft/s$ to $45~ft/s$, the distance increases from $64.4 ~ft$ to $69.9 ~ft$. As $v$ increases, the distance increases significantly.