Question

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Part 1
An arrow is fired into the air with an initial velocity of 128 feet per second. The height in feet of the arrow t seconds after it was shot into the air is given by the function h left parenthesis x right parenthesis equals negative 16 t squared plus 128 t. Find the maximum height of the arrow.
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Part 1
A.
64 ft
B.
448 ft
C.
256 ft
D.
768 ft

Answer 1

To find the maximum height of the arrow, we can analyze the height function given by the formula:

\[
h(t) = -16t^2 + 128t
\]

This is a quadratic function in the standard form \(h(t) = at^2 + bt + c\), where \(a = -16\), \(b = 128\), and \(c = 0\). The maximum value of a quadratic function occurs at the vertex, which can be found using the formula:

\[
t = -\frac{b}{2a}
\]

Substituting the values of \(a\) and \(b\):

\[
t = -\frac{128}{2 \times -16} = \frac{128}{32} = 4
\]

Now, we substitute \(t = 4\) back into the height function to find the maximum height:

\[
h(4) = -16(4^2) + 128(4)
\]
\[
= -16(16) + 128(4)
\]
\[
= -256 + 512
\]
\[
= 256
\]

Thus, the maximum height of the arrow is **256 feet**.

The correct answer is **C. 256 ft**.