SOLUTION: The height h in feet of an object after t seconds is given by the function h = –16t^2 + 30t + 10. How long will it take the object to hit the ground? Round your answer to the ne

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Question 43206This question is from textbook Beginning Algebra
: The height h in feet of an object after t seconds is given by the function h = –16t^2 + 30t + 10. How long will it take the object to hit the ground? Round your answer to the nearest thousandth.

Umm....huh? Please help
This question is from textbook Beginning Algebra

Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
It depends, are you throwing the ball of a 10-foot building or are you throwing the ball onto the ground where you are 10 feet high.
If you are throwing the ball off a 10-foot building (which is likely the case):
We want the height to be zero.
h+=+0
h+=+-16t%5E2+%2B+30t+%2B+10
0+=+-16t%5E2+%2B+30t+%2B+10
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation at%5E2%2Bbt%2Bc=0 (in our case -16t%5E2%2B30t%2B10+=+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=%2830%29%5E2-4%2A-16%2A10=1540.

Discriminant d=1540 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-30%2B-sqrt%28+1540+%29%29%2F2%5Ca.

t%5B1%5D+=+%28-%2830%29%2Bsqrt%28+1540+%29%29%2F2%5C-16+=+-0.288838554396786
t%5B2%5D+=+%28-%2830%29-sqrt%28+1540+%29%29%2F2%5C-16+=+2.16383855439679

Quadratic expression -16t%5E2%2B30t%2B10 can be factored:
-16t%5E2%2B30t%2B10+=+-16%28t--0.288838554396786%29%2A%28t-2.16383855439679%29
Again, the answer is: -0.288838554396786, 2.16383855439679. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B30%2Ax%2B10+%29

About: 2.16 seconds