Question 91123: Hello, I need help with this question. The equation is h=-16t^2+112t this makes an equation of the height of an arrow shot upward with a velocity of 112ft/s t=the time the arrow leaves the ground. What is the time it takes for an arrow to reach a height of 180.
I got 180=-16t^2+112t
But I don't know where to go from here I am so confused.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! This same problem came up a few months ago. This is what I submitted then
:
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.
:
Substitute 180 for h in the given equation: -16t^2 + 112t = h
:
-16t^2 + 112t = 180
:
-16t^2 + 112t - 180 = 0; subtract 180 from both sides, gives us a quadratic eq:
:
Simplify divide equation by -4, that changes the signs and gives you:
4t^2 - 28t + 45 = 0
:
Factor this to:
(2t - 5)(2t - 9) = 0
:
2t = +5
t = 2.5 sec (on the way up)
and
2t = +9
t = 4.5 sec (on the way down)
:
:
Check solution using t = 2.5, in the original equation:
-16(2.5^2) + 112(2.5) =
-16(6.25) + 280 =
-100 + 280 = 180
:
You can check it using the t = 4.5 solution
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