Question

a person shoots an arrow vertically into the air from a height of 6 feet with an initial velocity of 96 feet per second. the height, h, in feet above the ground, at any time, t (in seconds), is modeled by h(t)=6 + 96t - 16t². a) determine the maximum height the arrow will attain. feet

Explanation:

Step1: Identify the vertex of the quadratic function

The height function is \( h(t) = -16t^2 + 96t + 6 \), which is a quadratic function in the form \( y = ax^2 + bx + c \) where \( a = -16 \), \( b = 96 \), and \( c = 6 \). For a quadratic function, the time \( t \) at which the maximum (since \( a < 0 \)) occurs is given by \( t = -\frac{b}{2a} \).

Substitute \( a = -16 \) and \( b = 96 \) into the formula:
\( t = -\frac{96}{2(-16)} = -\frac{96}{-32} = 3 \) seconds.

Step2: Find the maximum height by substituting \( t = 3 \) into \( h(t) \)

Now substitute \( t = 3 \) into the height function \( h(t) = 6 + 96t - 16t^2 \):

\( h(3) = 6 + 96(3) - 16(3)^2 \)

First, calculate the powers and multiplications:
\( 96(3) = 288 \)
\( 16(3)^2 = 16 \times 9 = 144 \)

Then substitute these values back into the equation:
\( h(3) = 6 + 288 - 144 \)

Now perform the additions and subtractions:
\( 6 + 288 = 294 \)
\( 294 - 144 = 150 \)

So the maximum height the arrow will attain is 150 feet.

Answer:

150