SOLUTION: A ball is thrown vertically upward from the top of a building 144 feet tall with an initial velocity of 128 feet per second. The distance s (in feet) from the ground after t secon

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Question 159973: A ball is thrown vertically upward from the top of a building 144 feet tall with an initial velocity of 128 feet per second. The distance s (in feet) from the ground after t seconds is s = 144 + 128t -16t2. After how many seconds will the ball pass the top of the building on the way down?
Answer by KnightOwlTutor(293) About Me  (Show Source):
You can put this solution on YOUR website!
This is a quadratic equation. We want to know how long it takes to reach the maximum height of 144 ft. This is an upside down parabola.
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -16x%5E2%2B128x%2B144+=+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=%28128%29%5E2-4%2A-16%2A144=25600.

Discriminant d=25600 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-128%2B-sqrt%28+25600+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28128%29%2Bsqrt%28+25600+%29%29%2F2%5C-16+=+-1
x%5B2%5D+=+%28-%28128%29-sqrt%28+25600+%29%29%2F2%5C-16+=+9

Quadratic expression -16x%5E2%2B128x%2B144 can be factored:
-16x%5E2%2B128x%2B144+=+%28x--1%29%2A%28x-9%29
Again, the answer is: -1, 9. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B128%2Ax%2B144+%29

Since time cannot be negative the answer is 9s