SOLUTION: s=vt+16t2 object released from 684 feet the initial velocity is 18 ft/s and air resistance is neglected how many seconds will the object hit the ground?

Algebra ->  Systems-of-equations -> SOLUTION: s=vt+16t2 object released from 684 feet the initial velocity is 18 ft/s and air resistance is neglected how many seconds will the object hit the ground?      Log On


   

Question 926001: s=vt+16t2
object released from 684 feet the initial velocity is 18 ft/s and air resistance is neglected how many seconds will the object hit the ground?
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
s=vt+16t2
object released from 684 feet the initial velocity is 18 ft/s and air resistance is neglected how many seconds will the object hit the ground?
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Use s(t) = -16t^2 + vt + so
If the in+tial velocity is downward (you didn't say) it's -18. If it's upward, it's +18.
I'll assume downward.
so is the starting height.
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s(t) = -16t^2 - 18t + 684 = 0
Solve for t
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -16x%5E2%2B-18x%2B684+=+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=%28-18%29%5E2-4%2A-16%2A684=44100.

Discriminant d=44100 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--18%2B-sqrt%28+44100+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-18%29%2Bsqrt%28+44100+%29%29%2F2%5C-16+=+-7.125
x%5B2%5D+=+%28-%28-18%29-sqrt%28+44100+%29%29%2F2%5C-16+=+6

Quadratic expression -16x%5E2%2B-18x%2B684 can be factored:
-16x%5E2%2B-18x%2B684+=+%28x--7.125%29%2A%28x-6%29
Again, the answer is: -7.125, 6. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B-18%2Ax%2B684+%29

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Ignore the negative solution.
t = 6 seconds