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July 30, 2014

July 30, 2014

Posted by **Kai** on Monday, February 28, 2011 at 9:25pm.

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?

- Algebra 1 -
**Helper**, Monday, February 28, 2011 at 10:31pmy = 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.

- Algebra 1 -
**Kai**, Monday, February 28, 2011 at 11:00pmThank you so much, this really helped! :)

- Algebra 1 -
**Helper**, Monday, February 28, 2011 at 11:02pmYou're welcome :)

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