SOLUTION: An object is projected upward from the top of a tower. Its distance in feet above the ground after t seconds is given by s(t)=-16t^2+64t+80. How many seconds will it take to reach

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Question 7496: An object is projected upward from the top of a tower. Its distance
in feet above the ground after t seconds is given by s(t)=-16t^2+64t+80. How many seconds will it take to reach ground
level?
Answer by ichudov(507) About Me  (Show Source):
You can put this solution on YOUR website!
all you have to do is notice that when it hits the ground, its s(t) would be zero (zero height).
so you say
-16t^2+64t+80=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation at%5E2%2Bbt%2Bc=0 (in our case -16t%5E2%2B64t%2B80+=+0) has the following solutons:

t%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=%2864%29%5E2-4%2A-16%2A80=9216.

Discriminant d=9216 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-64%2B-sqrt%28+9216+%29%29%2F2%5Ca.

t%5B1%5D+=+%28-%2864%29%2Bsqrt%28+9216+%29%29%2F2%5C-16+=+-1
t%5B2%5D+=+%28-%2864%29-sqrt%28+9216+%29%29%2F2%5C-16+=+5

Quadratic expression -16t%5E2%2B64t%2B80 can be factored:
-16t%5E2%2B64t%2B80+=+-16%28t--1%29%2A%28t-5%29
Again, the answer is: -1, 5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B64%2Ax%2B80+%29


t=-1 is a wrong answer, so 5 is the right one.