SOLUTION: A projectile is launched into the air with an initial upward velocity of 24 feet per second. The height (h) in feet of the projectile can be modeled by h = -16t^2 + 24t + 3, where

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A projectile is launched into the air with an initial upward velocity of 24 feet per second. The height (h) in feet of the projectile can be modeled by h = -16t^2 + 24t + 3, where      Log On


   

Question 467061: A projectile is launched into the air with an initial upward velocity of 24 feet per second. The height (h) in feet of the projectile can be modeled by h = -16t^2 + 24t + 3, where t is the time in seconds after it was launched. How much time, in seconds, does it take the object to reach 12 feet above the ground?
( TOPIC: Quadratic Equations )
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
h(t) = -16t^2 + 24t + 3
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12 = -16t^2 + 24t + 3
-16t^2 + 24t - 9 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -16x%5E2%2B24x%2B-9+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2824%29%5E2-4%2A-16%2A-9=0.

Discriminant d=0 is zero! That means that there is only one solution: x+=+%28-%2824%29%29%2F2%5C-16.
Expression can be factored: -16x%5E2%2B24x%2B-9+=+%28x-0.75%29%2A%28x-0.75%29

Again, the answer is: 0.75, 0.75. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B24%2Ax%2B-9+%29

t = x
t = 0.75 seconds