SOLUTION: I need help solving this problem. an arrow is shot into the sky with initial velocity 100ft/sec from a height of 180 feet above the ground. The function h(t) = -16t^2 +100t +18

Question 734179: I need help solving this problem.
an arrow is shot into the sky with initial velocity 100ft/sec from a height of 180 feet above the ground. The function h(t) = -16t^2 +100t +180 models the arrow's height (h)(t), in feet) as a function of time (t, in seconds).
1. After how many seconds does the arrow reach its highest point? Round to 4 decimals.
I graphed this equation and tried to find the vertex and came up with 3.2447 seconds, which isn't right.
That set me up to fail the second part which is to figure out how high the arrow is at its highest point.
I substituted 3.2447 into the equation to get 225.484 feet and that was wrong too.
So then I couldn't get the practical range of the function either.
I was able to solve how many seconds it took for the arrow to hit the ground (7.709) and the practical domain, but I can't solve the range.
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
The function h(t) = -16t^2 +100t +180 models the arrow's height (h)(t), in feet) as a function of time (t, in seconds).
1. After how many seconds does the arrow reach its highest point? Round to 4 decimals.
.
The equation:
h(t) = -16t^2 +100t +180
is a "quadratic" or parabola because it is a polynomial of degree 2.
We also know that it is opened downwards because the leading coefficient is negative. Therefore, the "axis of symmetry" will be the time that the arrow is at its highest point:
t = -b/(2a)
t = -100/(2(-16))
t = -100/(-32)
t = 3.1250 seconds
.
2. how high the arrow is at its highest point
Plug above back into the equation to find highest point:
h(3.125) = -16(3.125)^2 +100(3.125) +180
h(3.125) = 336.2500 feet

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