SOLUTION: An object is thrown upward from the top of an 80 foot building with an initial velocity of 64 feet per second. The height h of an object after t seconds is given by the quadratic e

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Question 211479This question is from textbook Beginning & Intermediate Algebra
: An object is thrown upward from the top of an 80 foot building with an initial velocity of 64 feet per second. The height h of an object after t seconds is given by the quadratic equation (((h=-16t^2+64t+80))). When will the object hit the ground. This question is from textbook Beginning & Intermediate Algebra

Found 2 solutions by nerdybill, MathTherapy:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
The problem gives you:
h=-16t^2+64t+80
where
h is height in feet
t is time in secs
.
If it hits the ground, h = 0 so
set h=0 and solve for t:
h=-16t^2+64t+80
0=-16t^2+64t+80
0=-4t^2+16t+5
Solving using the quadratic equation yields:
t = {-0.29, 4.29}
We can throw out the negative solution leaving
t = 4.29 seconds
.
Details of quadratic to follow:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation at%5E2%2Bbt%2Bc=0 (in our case -4t%5E2%2B16t%2B5+=+0) has the following solutons:

t%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=%2816%29%5E2-4%2A-4%2A5=336.

Discriminant d=336 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-16%2B-sqrt%28+336+%29%29%2F2%5Ca.

t%5B1%5D+=+%28-%2816%29%2Bsqrt%28+336+%29%29%2F2%5C-4+=+-0.29128784747792
t%5B2%5D+=+%28-%2816%29-sqrt%28+336+%29%29%2F2%5C-4+=+4.29128784747792

Quadratic expression -4t%5E2%2B16t%2B5 can be factored:
-4t%5E2%2B16t%2B5+=+-4%28t--0.29128784747792%29%2A%28t-4.29128784747792%29
Again, the answer is: -0.29128784747792, 4.29128784747792. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-4%2Ax%5E2%2B16%2Ax%2B5+%29

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
An object is thrown upward from the top of an 80 foot building with an initial velocity of 64 feet per second. The height h of an object after t seconds is given by the quadratic equation (((h=-16t^2+64t+80))). When will the object hit the ground.

Since we're looking for "h," or the height at 0, we then set the equation to 0, and solve for t. We will then have:
0+=+-16t%5E2+%2B+64t+%2B+80
-16%280%29+=+-16%28t%5E2+-+4t+-+5%29 ------> factoring out the GCF, -16
0+=+t%5E2+-+4t+-+5 -------> -16+=+-16
0 = (t - 5)(t + 1)

Therefore, t = 5, or -1, but t = - 1 is ignored
This means that the object will hit the ground highlight_green%285%29 seconds after being thrown in the air.