SOLUTION: when a ball is thrown, it's height in feet H after t seconds is given by the equation h=vt-16t^2 where v is the initial upwards velocity in feet per second if v=13 feet per second

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Question 330347: when a ball is thrown, it's height in feet H after t seconds is given by the equation h=vt-16t^2 where v is the initial upwards velocity in feet per second if v=13 feet per second find all the values of t for which h=2
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
when a ball is thrown, it's [sic] height in feet H after t seconds is given by the equation h=vt-16t^2 where v is the initial upwards velocity in feet per second if v=13 feet per second find all the values of t for which h=2
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h(t) = vt-16t^2
Sub 2 for h and 13 for h
2 = 13t - 16t^2
16t^2 - 13t + 2 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -16x%5E2%2B13x%2B-2+=+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=%2813%29%5E2-4%2A-16%2A-2=41.

Discriminant d=41 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-13%2B-sqrt%28+41+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2813%29%2Bsqrt%28+41+%29%29%2F2%5C-16+=+0.206152367580223
x%5B2%5D+=+%28-%2813%29-sqrt%28+41+%29%29%2F2%5C-16+=+0.606347632419777

Quadratic expression -16x%5E2%2B13x%2B-2 can be factored:
-16x%5E2%2B13x%2B-2+=+%28x-0.206152367580223%29%2A%28x-0.606347632419777%29
Again, the answer is: 0.206152367580223, 0.606347632419777. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B13%2Ax%2B-2+%29

t = x
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It's at 2 feet twice, going up, then coming back down.
Most people find it difficult to throw a ball straight up from ground level.