SOLUTION: an arrow is shot vertically from a platform 22 ft high at a rate of 175 ft per second. when will the arrow hit the ground

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Question 813566: an arrow is shot vertically from a platform 22 ft high at a rate of 175 ft per second. when will the arrow hit the ground
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
an arrow is shot vertically from a platform 22 ft high at a rate of 175 ft per second. when will the arrow hit the ground
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You didn't spec a function.
Most commonly used on Earth (in feet) is;
h(t) = -16t^2 + 175t + 22 for the given initial conditions.
h(t) = -16t^2 + 175t + 22 = 0
Solve for t, use the positive solution.
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -16x%5E2%2B175x%2B22+=+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=%28175%29%5E2-4%2A-16%2A22=32033.

Discriminant d=32033 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-175%2B-sqrt%28+32033+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28175%29%2Bsqrt%28+32033+%29%29%2F2%5C-16+=+-0.124301632382808
x%5B2%5D+=+%28-%28175%29-sqrt%28+32033+%29%29%2F2%5C-16+=+11.0618016323828

Quadratic expression -16x%5E2%2B175x%2B22 can be factored:
-16x%5E2%2B175x%2B22+=+%28x--0.124301632382808%29%2A%28x-11.0618016323828%29
Again, the answer is: -0.124301632382808, 11.0618016323828. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B175%2Ax%2B22+%29

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t =~ 11.062 seconds