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 **
<u>Constant Acceleration Equations</u> (in ft)
- s = displacement in ft
- u = initial velocity in ft/s
- v = final velocity in ft/s
- a = acceleration in ft/s²
- t = time in s (seconds)
Assuming that the projectile is fired at an angle of 12° <u>above the horizontal.</u>
Consider the vertical and horizontal motion separately.
<u>Vertical motion</u>
First, find time by resolving vertically, taking up as positive and
g = 32 ft/s²
- s = 0
- u = 217 sin 12°
- v =
- a = -32
- t = t
<u>Horizontal motion</u>
Resolving horizontally, taking right as positive:
(The horizontal component of velocity is constant, as there is no acceleration horizontally)
- s = s
- u = 217 cos 12°
- v = 217 cos 12°
- a = 0
- t = 2.819802307...
Therefore, the horizontal distance is 599 ft (nearest foot)