Question

if a baseball is projected upward from ground level with an initial velocity of 96 feet per second, then its height is a function of time, given by s = - 16t² + 96t. what is the maximum height reached by the ball? the maximum height reached by the ball is □ feet.

Explanation:

Step1: Identify the function form

The height function \(s = - 16t^{2}+96t\) is a quadratic function of the form \(y = ax^{2}+bx + c\), where \(a=-16\), \(b = 96\), \(c = 0\).

Step2: Find the time \(t\) at which the maximum occurs

For a quadratic function \(y=ax^{2}+bx + c\), the \(x\) - coordinate (in this case \(t\)) of the vertex is given by \(t=-\frac{b}{2a}\).
Substitute \(a=-16\) and \(b = 96\) into the formula:
\(t=-\frac{96}{2\times(-16)}=\frac{-96}{-32}=3\)

Step3: Find the maximum height

Substitute \(t = 3\) into the height function \(s=-16t^{2}+96t\)
\(s=-16\times3^{2}+96\times3\)
\(s=-16\times9 + 288\)
\(s=-144+288\)
\(s = 144\)

Answer:

144