To determine after how many seconds the projectile will reach a height of 200 feet, we need to use the equation of motion:
[tex]\[ h(t) = at^2 + vt + h_0 \][/tex]
where:
- [tex]\( h(t) \)[/tex] is the height of the projectile at time [tex]\( t \)[/tex],
- [tex]\( a \)[/tex] is the acceleration due to gravity, which is [tex]\(-16 \, ft/s^2 \)[/tex] (negative because gravity acts downward),
- [tex]\( v \)[/tex] is the initial velocity, [tex]\(120 \, ft/s \)[/tex],
- [tex]\( h_0 \)[/tex] is the initial height, [tex]\(0 \, ft \)[/tex] (since it is launched from the ground).
We need to find the times [tex]\( t \)[/tex] when the height [tex]\( h(t) \)[/tex] is 200 feet. Plugging all known values into the equation, we get:
[tex]\[ 200 = -16t^2 + 120t + 0 \][/tex]
This simplifies to:
[tex]\[ -16t^2 + 120t - 200 = 0 \][/tex]
This is a quadratic equation in the form [tex]\( at^2 + bt + c = 0 \)[/tex], where:
- [tex]\( a = -16 \)[/tex],
- [tex]\( b = 120 \)[/tex],
- [tex]\( c = -200 \)[/tex].
To solve for [tex]\( t \)[/tex], we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula:
[tex]\[ t = \frac{-120 \pm \sqrt{(120)^2 - 4(-16)(-200)}}{2(-16)} \][/tex]
[tex]\[ t = \frac{-120 \pm \sqrt{14400 - 12800}}{-32} \][/tex]
[tex]\[ t = \frac{-120 \pm \sqrt{1600}}{-32} \][/tex]
[tex]\[ t = \frac{-120 \pm 40}{-32} \][/tex]
This gives us two solutions for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-120 + 40}{-32} = \frac{-80}{-32} = 2.5 \, s \][/tex]
[tex]\[ t = \frac{-120 - 40}{-32} = \frac{-160}{-32} = 5.0 \, s \][/tex]
Therefore, the projectile will reach a height of 200 feet at approximately [tex]\( 2.5 \, s \)[/tex] and [tex]\( 5.0 \, s \)[/tex]. Among the given options, [tex]\( 2.5 \, s \)[/tex] is a correct answer.
Hence, the correct answer is:
[tex]\[ \boxed{2.5 \, s} \][/tex]
