# Find k

• Jan 18th 2007, 06:33 PM
symmetry
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
• Jan 18th 2007, 06:37 PM
topsquark
Quote:

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:
$h'(t) = -32t + k = 0$

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

-Dan
• Jan 18th 2007, 06:41 PM
symmetry
ok
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 PM
topsquark
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 $y = ax^2 + bx + c$ the axis of symmetry will be the line $x = -\frac{b}{2a}$.

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

So
$k = 128$.

-Dan
• Jan 20th 2007, 04:56 AM
symmetry
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!
• Jan 20th 2007, 09:32 AM
CaptainBlack
Quote:

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