Question

GIVING BRAINLIEST AND MAX POINTS! ANSWER ASAPPP! The initial velocity of a projectile is 217 ft/sec, and the firing angle is 12°. Calculate the horizontal distance.
a
599 feet
b
699 feet
c
499 feet
d
799 feet

Answer 1

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)

• 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° 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

• s = 0
• u = 217 sin 12°
• v =
• a = -32
• t = t

$\begin{aligned}\textsf{Using}\quad s & =ut+\dfrac{1}{2}at^2\\\implies 0 & = 217 \sin 12^{\circ}t+\dfrac{1}{2}(-32)t^2\\0 & = 217 \sin 12^{\circ}t-16t^2\\t(16t-217 \sin 12^{\circ}) & = 0\\16t & = 217 \sin 12^{\circ}\\t & = \dfrac{217 \sin 12^{\circ}}{16}\\t & = 2.819802307...\: \sf s\end{aligned}$

Horizontal motion

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...

$\begin{aligned}\textsf{Using}\quad s & =ut+\dfrac{1}{2}at^2\\\implies s & = (217 \cos 12^{\circ})(2.819802307...)+\dfrac{1}{2}(0)t^2\\& =598.5256808...\end{aligned}$

Therefore, the horizontal distance is 599 ft (nearest foot)