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

Algebra ->  Points-lines-and-rays -> SOLUTION: The equation h= -16t^2+112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112 ft/s, where t is the time after the arrow leaves the ground. F      Log On


   

Question 125190: The equation h= -16t^2+112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112 ft/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.
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
h=+-16t%5E2%2B112t is your equation for height. Since you are given the height of 180 ft., just replace h with 180 and solve it like any other quadratic equation.

-16t%5E2%2B112t=180

-16t%5E2%2B112t-180=0

If you divide through by -4, the coefficients become a little more manageable:

4t%5E2-28t%2B45=0

You can use the quadratic formula if you don't want to scratch your head over the factorization of this one, but the factors come out to:

%282x-9%29%282x-5%29=0

So, x=9%2F2 or x=5%2F2

Therefore, the arrow will be 180ft above the ground 2.5 seconds after it is released on the way up, and again at 4.5 seconds on the way down.