SOLUTION: an object is dropped from 38 feet below the tip of the pinnacle atop a 1482-ft tall building. the height h of the object after t seconds is given by the equation h=-16t^2+1444. fin

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Question 949111: an object is dropped from 38 feet below the tip of the pinnacle atop a 1482-ft tall building. the height h of the object after t seconds is given by the equation h=-16t^2+1444. find how many seconds pass before the object reaches the ground.
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
The point when it will hit the ground is when h=0
-16t^2+1444=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation at%5E2%2Bbt%2Bc=0 (in our case -16t%5E2%2B0t%2B1444+=+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=%280%29%5E2-4%2A-16%2A1444=92416.

Discriminant d=92416 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-0%2B-sqrt%28+92416+%29%29%2F2%5Ca.

t%5B1%5D+=+%28-%280%29%2Bsqrt%28+92416+%29%29%2F2%5C-16+=+-9.5
t%5B2%5D+=+%28-%280%29-sqrt%28+92416+%29%29%2F2%5C-16+=+9.5

Quadratic expression -16t%5E2%2B0t%2B1444 can be factored:
-16t%5E2%2B0t%2B1444+=+-16%28t--9.5%29%2A%28t-9.5%29
Again, the answer is: -9.5, 9.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B0%2Ax%2B1444+%29

The answer is 9.5 seconds
CHECK
0=-16t^2+1444
0=-16(9.5)^2 + 1444
0=-16(90.25)+1444
0==1444+1444
0=0