SOLUTION: the height h(t), in feet, of an airborne tee-shirt t seconds after being launched can be approximated by h(t)= -15t^2+105t+10,0≤t≤10 find the times when the tee-shirt w

Algebra ->  Rectangles -> SOLUTION: the height h(t), in feet, of an airborne tee-shirt t seconds after being launched can be approximated by h(t)= -15t^2+105t+10,0≤t≤10 find the times when the tee-shirt w      Log On


   

Question 921914: the height h(t), in feet, of an airborne tee-shirt t seconds after being launched can be approximated by h(t)= -15t^2+105t+10,0≤t≤10 find the times when the tee-shirt will reach a fan 160 feet above ground level.
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
h(t)= -15t^2+105t+10,0≤t≤10 find the times when the tee-shirt will reach a fan 160 feet above ground level.
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h(t)= -15t^2+105t+10 = 160
-3t^2 + 21t + 2 = 32
-3t^2 + 21t - 30 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -3x%5E2%2B21x%2B-30+=+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=%2821%29%5E2-4%2A-3%2A-30=81.

Discriminant d=81 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-21%2B-sqrt%28+81+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2821%29%2Bsqrt%28+81+%29%29%2F2%5C-3+=+2
x%5B2%5D+=+%28-%2821%29-sqrt%28+81+%29%29%2F2%5C-3+=+5

Quadratic expression -3x%5E2%2B21x%2B-30 can be factored:
-3x%5E2%2B21x%2B-30+=+%28x-2%29%2A%28x-5%29
Again, the answer is: 2, 5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-3%2Ax%5E2%2B21%2Ax%2B-30+%29

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t = 2 seconds going up
t = 5 second coming down