document.write( "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 \n" ); document.write( "
Algebra.Com's Answer #489772 by Alan3354(69443)\"\" \"About 
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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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\n" ); document.write( "You didn't spec a function.
\n" ); document.write( "Most commonly used on Earth (in feet) is;
\n" ); document.write( "h(t) = -16t^2 + 175t + 22 for the given initial conditions.
\n" ); document.write( "h(t) = -16t^2 + 175t + 22 = 0
\n" ); document.write( "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:
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\n" ); document.write( " \"x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"
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\n" ); document.write( " For these solutions to exist, the discriminant \"b%5E2-4ac\" should not be a negative number.
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\n" ); document.write( " First, we need to compute the discriminant \"b%5E2-4ac\": \"b%5E2-4ac=%28175%29%5E2-4%2A-16%2A22=32033\".
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\n" ); document.write( " 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\".
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\n" ); document.write( " \"x%5B1%5D+=+%28-%28175%29%2Bsqrt%28+32033+%29%29%2F2%5C-16+=+-0.124301632382808\"
\n" ); document.write( " \"x%5B2%5D+=+%28-%28175%29-sqrt%28+32033+%29%29%2F2%5C-16+=+11.0618016323828\"
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\n" ); document.write( " Quadratic expression \"-16x%5E2%2B175x%2B22\" can be factored:
\n" ); document.write( " \"-16x%5E2%2B175x%2B22+=+%28x--0.124301632382808%29%2A%28x-11.0618016323828%29\"
\n" ); document.write( " Again, the answer is: -0.124301632382808, 11.0618016323828.\n" ); document.write( "Here's your graph:
\n" ); document.write( "\"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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\n" ); document.write( "t =~ 11.062 seconds
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