Answer:
a) 599 feet
Step-by-step explanation:
** As the measure of distance is in feet, we must use the value of gravity in ft/s² **
Constant Acceleration Equations (in ft)
Assuming that the projectile is fired at an angle of 12° above the horizontal.
Consider the vertical and horizontal motion separately.
Vertical motion
First, find time by resolving vertically, taking up as positive and
g = 32 ft/s²
[tex]\begin{aligned}\textsf{Using}\quad s & =ut+\dfrac{1}{2}at^2\\\implies 0 & = 217 \sin 12^{\circ}t+\dfrac{1}{2}(-32)t^2\\0 & = 217 \sin 12^{\circ}t-16t^2\\t(16t-217 \sin 12^{\circ}) & = 0\\16t & = 217 \sin 12^{\circ}\\t & = \dfrac{217 \sin 12^{\circ}}{16}\\t & = 2.819802307...\: \sf s\end{aligned}[/tex]
Horizontal motion
Resolving horizontally, taking right as positive:
(The horizontal component of velocity is constant, as there is no acceleration horizontally)
[tex]\begin{aligned}\textsf{Using}\quad s & =ut+\dfrac{1}{2}at^2\\\implies s & = (217 \cos 12^{\circ})(2.819802307...)+\dfrac{1}{2}(0)t^2\\& =598.5256808...\end{aligned}[/tex]
Therefore, the horizontal distance is 599 ft (nearest foot)
