SOLUTION: The equation h= -16t^2 + 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.
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Question 65745: The equation h= -16t^2 + 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 180ft.
h = 16t^2 + 112t
H= 180ft
180ft = -16t^2 + 112t
From here I seem to be lost on how to solve the problem. It tells me the information that I need to know I guess I just do not know how to use it... thank you for the help. Answer by Nate(3500) (Show Source):
You can put this solution on YOUR website! 180 = -16t^2 + 112t
0 = -16^2 + 112t - 180
From here, you could use quadratic formula:
for ~> ax^2 + bx + c
where:
a = -16
b = 112
c = -180
Or, you could just use solving cubes....
180 = -16t^2 + 112t
-11.25 = t^2 - 7t
1 = (t - 3.5)^2
3.5 +- 1 = t
t = 4.5 seconds and t = 2.5 seconds
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