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)   (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 (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=92416 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: -9.5, 9.5. Here's your graph:

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

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