SOLUTION: The equation h = -16t^2 – 32t + 112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112ft/s, where t is the time after the arrow leaves t

Algebra ->  Expressions-with-variables -> SOLUTION: The equation h = -16t^2 – 32t + 112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112ft/s, where t is the time after the arrow leaves t      Log On


   

Question 45912: The equation h = -16t^2 – 32t + 112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112ft/s, where t is the time after the arrow leaves the ground. Find the time it takes for the arrow to reach a height of 180 ft.
Not sure where to start on this one.
Thanks for help
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
I do not know how you get -32t ....
h+=+-16t%5E2+%96+32t+%2B+112t
180+=+-16t%5E2+%2B+80t
0+=+-4t%5E2+%2B+20t+-+45
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation at%5E2%2Bbt%2Bc=0 (in our case -4t%5E2%2B20t%2B-45+=+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=%2820%29%5E2-4%2A-4%2A-45=-320.

The discriminant -320 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -320 is + or - sqrt%28+320%29+=+17.8885438199983.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-4%2Ax%5E2%2B20%2Ax%2B-45+%29

h+=+-16t%5E2+%2B+80t
V(2.5,100)
The highest the arrow will go is 100ft.