1. ## Find 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

2. Originally Posted by symmetry
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
Max height is when the derivative of the height function goes to 0. So:
$\displaystyle h'(t) = -32t + k = 0$

Thus $\displaystyle k = 32t = 32 \cdot 4 = 128$

-Dan

3. ## ok

Derivative is a calculus word.

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

Thanks anyway!

4. 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

5. ## ok

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!

6. Originally Posted by symmetry
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!
I think we would all be interested to know what test you are preparing for.
Its always nice to have some context when subjected to an avalache of
questions

RonL