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

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) (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 (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=25600 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -1, 9. Here's your graph:

Since time cannot be negative the answer is 9s

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