An archer shoots an arrow into the air such that its height at any time, t, is given by the function h(t) = -16t^2 + kt + 3. If the maximum height of the arrow occurs at time t = 4, what is the value of k?

(1) 128 (3) 8

(2) 64 (4) 4

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- Jan 18th 2007, 06:33 PMsymmetryFind k
An archer shoots an arrow into the air such that its height at any time, t, is given by the function h(t) = -16t^2 + kt + 3. If the maximum height of the arrow occurs at time t = 4, what is the value of k?

(1) 128 (3) 8

(2) 64 (4) 4 - Jan 18th 2007, 06:37 PMtopsquark
- Jan 18th 2007, 06:41 PMsymmetryok
Derivative is a calculus word.

I am not asking calculus questions at this point in the study book.

Thanks anyway! - Jan 18th 2007, 06:54 PMtopsquark
All right. In that case we need to look at the height function a bit more carefully. It is an inverted parabola. The maximum height will be at the vertex of the parabola, which is on the axis of symmetry.

Given a parabola $\displaystyle y = ax^2 + bx + c$ the axis of symmetry will be the line $\displaystyle x = -\frac{b}{2a}$.

We have the parabola $\displaystyle h = -16t^2 + kt + 3$. We know that the location of the max height is the line t = 4, which is our axis of symmetry. So:

$\displaystyle t = - \frac{k}{2 \cdot -16} = 4$

So

$\displaystyle k = 128$.

-Dan - Jan 20th 2007, 04:56 AMsymmetryok
Dan,

Thanks for breaking the question some other way.

I am not ready for calculus questions at this point in the course of my review for the June state test.

Thanks! - Jan 20th 2007, 09:32 AMCaptainBlack