Asked by Kai
Question:
The height in feet above the ground of an arrow t seconds after it is shot can be modeled by y = -16t(squared) + 63t + 4. Can the arrow pass over a tree that is 68 feet tall? Explain.
Completely lost, could anyone help me out?
The height in feet above the ground of an arrow t seconds after it is shot can be modeled by y = -16t(squared) + 63t + 4. Can the arrow pass over a tree that is 68 feet tall? Explain.
Completely lost, could anyone help me out?
Answers
Answered by
Helper
y = height = -16t^2 + 63t + 4
The vertex of this parabola is called the maximum point.
Vertex = -b/2a
Vertex = -63/(2(-16))
Vertex = t value = 1.96875
So, the maximum value for t seconds is 1.96875
y = height = -16t^2 + 63t + 4
t = 1.96875
y = -16(1.96875)^2 + 63(1.96875) + 4
y = -16(3.875977) + 124.0313 + 4
y = 66.0157
So, the maximum height = 66.0157 feet
Use could also use derivatives to find the maximum value for t.
The vertex of this parabola is called the maximum point.
Vertex = -b/2a
Vertex = -63/(2(-16))
Vertex = t value = 1.96875
So, the maximum value for t seconds is 1.96875
y = height = -16t^2 + 63t + 4
t = 1.96875
y = -16(1.96875)^2 + 63(1.96875) + 4
y = -16(3.875977) + 124.0313 + 4
y = 66.0157
So, the maximum height = 66.0157 feet
Use could also use derivatives to find the maximum value for t.
Answered by
Kai
Thank you so much, this really helped! :)
Answered by
Helper
You're welcome :)
Answered by
kkkkkkmmmmmm
dont do it
Answered by
Kiara
Thank you sooooo much. Literally been working on this for over an hour. LIFE SAVERRR
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