SOLUTION: An object is thrown downward from a plane which is 540 feet above the ground. The object is thrown with an initial velocity of 27 ft/sec. The distance traveled by the object in t s

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Question 311559: An object is thrown downward from a plane which is 540 feet above the ground. The object is thrown with an initial velocity of 27 ft/sec. The distance traveled by the object in t seconds is s = 16t2 + 27t, where s is in feet. How long does it take the object to reach the ground? Show your work.
I need help i understand that
540 = 16t^2 + 27t but what is after this I am lost. Please help me if you can
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
540 = 16t^2 + 27t
Considering up to be positive, it's
h(t) = -16t^2 - 27t + 540 = 0
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=35289 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: -6.71417707666998, 5.02667707666998. Here's your graph:
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