SOLUTION: A ball is thrown upward with a velociity of 80ft/s from a height of 576ft. H= -16t^2+80t=576 gives the height,h, after t sec. Find the time when the ball is 300ft above the groun

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A ball is thrown upward with a velociity of 80ft/s from a height of 576ft. H= -16t^2+80t=576 gives the height,h, after t sec. Find the time when the ball is 300ft above the groun      Log On


   

Question 605863: A ball is thrown upward with a velociity of 80ft/s from a height of 576ft. H= -16t^2+80t=576 gives the height,h, after t sec. Find the time when the ball is 300ft above the ground. H value is 300

OK--I know how that I have to set it to zero by subtracting 300 from 576. It will not factor out evenly. After setting it up with the quadratic formula, I am still getting it wrong. Please help me. I am on my 5th problem like this and am severly struggleing.
Found 2 solutions by jim_thompson5910, Alan3354:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
H=+-16t%5E2%2B80t%2B576

300+=+-16t%5E2%2B80t%2B576

0+=+-16t%5E2%2B80t%2B576-300

0+=+-16t%5E2%2B80t%2B276

-16t%5E2%2B80t%2B276+=+0

16t%5E2-80t-276+=+0

Now use the quadratic formula to solve

t+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

t+=+%28-%28-80%29%2B-sqrt%28%28-80%29%5E2-4%2816%29%28-276%29%29%29%2F%282%2816%29%29

t+=+%2880%2B-sqrt%286400-%28-17664%29%29%29%2F%2832%29

t+=+%2880%2B-sqrt%2824064%29%29%2F32

t+=+%2880%2Bsqrt%2824064%29%29%2F32 or t+=+%2880-sqrt%2824064%29%29%2F32

t+=+%2880%2B16%2Asqrt%2894%29%29%2F32 or t+=+%2880-16%2Asqrt%2894%29%29%2F32

t+=+%285%2Bsqrt%2894%29%29%2F2 or t+=+%285-sqrt%2894%29%29%2F2 <--- Exact Solutions

t+=+7.34767985741632 or t+=+-2.34767985741632 <--- Approximate Solutions


We now ignore the negative solution since a negative time value doesn't make any sense.


So the ball is 300 ft above the ground at approximately 7.34767985741632 seconds

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A ball is thrown upward with a velociity of 80ft/s from a height of 576ft. H= -16t^2+80t=576 gives the height,h, after t sec. Find the time when the ball is 300ft above the ground.
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It's H= -16t^2+80t + 576
H = 300
-16t^2+80t + 576 = 300
-16t^2+80t + 276 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -16x%5E2%2B80x%2B276+=+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=%2880%29%5E2-4%2A-16%2A276=24064.

Discriminant d=24064 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-80%2B-sqrt%28+24064+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2880%29%2Bsqrt%28+24064+%29%29%2F2%5C-16+=+-2.34767985741633
x%5B2%5D+=+%28-%2880%29-sqrt%28+24064+%29%29%2F2%5C-16+=+7.34767985741633

Quadratic expression -16x%5E2%2B80x%2B276 can be factored:
-16x%5E2%2B80x%2B276+=+%28x--2.34767985741633%29%2A%28x-7.34767985741633%29
Again, the answer is: -2.34767985741633, 7.34767985741633. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B80%2Ax%2B276+%29

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t =~ 7.348 seconds
Ignore the negative time.
Ignore the "It can be factored." statement, too.