a projectile is launched straight up with an initial velocity of 112 feet per second. how long would it take to reach 180 feet?

d = Vo*t + 0.5g*t^2 = 180,

112t - 16t^2 - 180 = 0,
Divide both sides by 4:
28t - 4t^2 - 45 = 0,
-4t^2 + 28t - 45 = 0.
X = 2.5s. Use Quadratic Formula and use
the smaller of the 2 solutions.

To find out how long it would take for the projectile to reach a height of 180 feet, we can use the equations of motion for a projectile in freefall.

In this case, the projectile is launched straight up, so the initial velocity (v₀) is positive (+112 ft/s) and the acceleration due to gravity (g) is negative (-32 ft/s²).

The equation we will use is:
s = v₀t + (1/2)gt²
where:
s = final height (180 ft)
v₀ = initial velocity (112 ft/s)
g = acceleration due to gravity (-32 ft/s²)
t = time taken (unknown)

Substituting the given values into the equation:
180 = 112t + (1/2)(-32)t²

Simplifying the equation:
180 = 112t - 16t²

Rearranging the equation:
16t² - 112t + 180 = 0

At this point, we have a quadratic equation, which can be solved using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:

t = ( -b ± √(b² - 4ac) ) / (2a)

For our equation:
a = 16
b = -112
c = 180

Substituting these values into the quadratic formula:
t = ( -(-112) ± √((-112)² - 4(16)(180)) ) / (2(16))
t = ( 112 ± √(12544 - 11520) ) / 32
t = ( 112 ± √1024 ) / 32
t = ( 112 ± 32 ) / 32

This gives us two solutions for t:
t₁ = ( 112 + 32 ) / 32 = 4
t₂ = ( 112 - 32 ) / 32 = 3

Therefore, it would take approximately 3 or 4 seconds for the projectile to reach a height of 180 feet, depending on the direction (up or down) and the timing of its launch.