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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

1 answer

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.