SOLUTION: A person standing close to the edge on the top of a 270-foot building throws a baseball vertically upward. The quadratic equation h = -16 t^2 + 160 t + 270 models the ball's

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Question 992194: A person standing close to the edge on the top of a 270-foot building throws a baseball vertically upward. The quadratic equation
h = -16 t^2 + 160 t + 270
models the ball's height \, h \, above the ground in feet, t seconds after it was thrown.
How high is the ball after 5 seconds?


How many seconds does it take until the ball finally hits the ground? Round to the nearest tenth of a second.
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
A person standing close to the edge on the top of a 270-foot building throws a baseball vertically upward. The quadratic equation
h = -16 t^2 + 160 t + 270
models the ball's height h above the ground in feet, t seconds after it was thrown.
How high is the ball after 5 seconds?
----
h(t) = -16t^2 + 160t + 270
Find h(5) = sub 5 for t.
h(5) is 670 feet.
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How many seconds does it take until the ball finally hits the ground? Round to the nearest tenth of a second.
h(t) = -16t^2 + 160t + 270
-16t^2 + 160t + 270 = 0
Solve for t. Use the positive solution.
-----
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=42880 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: -1.47108955277239, 11.4710895527724. Here's your graph:

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t = 11.5 seconds.
Maybe you didn't use the nearest 10th ?
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