To find the time it takes for a projectile launched straight up to reach a height of 200 feet, we can use the equation of motion for vertical displacement:
[tex]\[ h(t) = a t^2 + v t + h_0 \][/tex]
Where:
- [tex]\( h(t) \)[/tex] is the height at time [tex]\( t \)[/tex].
- [tex]\( a \)[/tex] is the acceleration due to gravity, given as [tex]\(-16 \, \text{ft/s}^2\)[/tex].
- [tex]\( v \)[/tex] is the initial velocity, given as [tex]\(120 \, \text{ft/s}\)[/tex].
- [tex]\( h_0 \)[/tex] is the initial height, which is [tex]\(0 \, \text{ft}\)[/tex] since the projectile is launched from the ground.
We are given that the height [tex]\( h(t) \)[/tex] is [tex]\(200 \, \text{ft}\)[/tex], and need to solve for [tex]\( t \)[/tex]:
[tex]\[ 200 = -16t^2 + 120t + 0 \][/tex]
This simplifies to the quadratic equation:
[tex]\[ -16t^2 + 120t - 200 = 0 \][/tex]
We can solve this quadratic equation in the standard form [tex]\( a t^2 + b t + c = 0 \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 120 \)[/tex]
- [tex]\( c = -200 \)[/tex]
Using the quadratic formula, [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
1. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 120^2 - 4 \cdot (-16) \cdot (-200) \][/tex]
[tex]\[ \Delta = 14400 - 12800 \][/tex]
[tex]\[ \Delta = 1600 \][/tex]
2. Substitute the discriminant back into the quadratic formula to find the possible values of [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-120 \pm \sqrt{1600}}{2 \cdot -16} \][/tex]
[tex]\[ t = \frac{-120 \pm 40}{-32} \][/tex]
This gives us two solutions:
[tex]\[ t_1 = \frac{-120 + 40}{-32} = \frac{-80}{-32} = 2.5 \][/tex]
[tex]\[ t_2 = \frac{-120 - 40}{-32} = \frac{-160}{-32} = 5.0 \][/tex]
Since time cannot be negative and we are looking for the moment when the object first reaches a height of 200 feet, the positive time value is:
[tex]\[ t = 2.5 \, \text{seconds} \][/tex]
Thus, the projectile will reach a height of 200 feet after approximately [tex]\(2.5\)[/tex] seconds.
Therefore, the correct answer is:
[tex]\[ \boxed{2.5 \, \text{seconds}} \][/tex]
