Question 97604
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.Solve to the nearest thousandth.
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h(t) = -16t^2+112t
Let h(t)= 180 ft and solve for "t":

-16t^2+112t=180
-16t^2+112t-180 = 0
Divide thru by -4 to get:
4t^2 -28t + 45 = 0
Use the quadratic formula to get:

t = [28 +- sqrt(28^2 - 4*4*45)]/8

t = [28 +- sqrt64)]/8

t = [28 +- 8]/8

t = (28+8)/8 = 4.5 seconds (time when the arrow is on the way down)
t = (28-8)/8 = 2.5 seconds (time when the arrow is on the way up)

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Cheers,
Stan H.