SOLUTION: An object is shot upward with an initial velocity of 240 feet per second so that its height s (in feet) above the ground after t seconds is given by s= -16t^2+240t. For what value

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Question 43041This question is from textbook algebra and trigonometry with analytic geometry
: An object is shot upward with an initial velocity of 240 feet per second so that its height s (in feet) above the ground after t seconds is given by s= -16t^2+240t. For what values of t will the object be at least 416 feet above the ground? I think I have it as (2,5). thanks for any help. This question is from textbook algebra and trigonometry with analytic geometry

Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
s+=+-16t%5E2+%2B+240t
416+%3C=+-16t%5E2+%2B+240t
0+%3C=+-16t%5E2+%2B+240t+-+416
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation at%5E2%2Bbt%2Bc=0 (in our case -16t%5E2%2B240t%2B-416+=+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=%28240%29%5E2-4%2A-16%2A-416=30976.

Discriminant d=30976 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-240%2B-sqrt%28+30976+%29%29%2F2%5Ca.

t%5B1%5D+=+%28-%28240%29%2Bsqrt%28+30976+%29%29%2F2%5C-16+=+2
t%5B2%5D+=+%28-%28240%29-sqrt%28+30976+%29%29%2F2%5C-16+=+13

Quadratic expression -16t%5E2%2B240t%2B-416 can be factored:
-16t%5E2%2B240t%2B-416+=+-16%28t-2%29%2A%28t-13%29
Again, the answer is: 2, 13. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B240%2Ax%2B-416+%29

Answer: 2%3C=t%3C=13