Question

if a ball is kicked straight up with an initial velocity of 96 ft/sec from a height of 4 ft, then its height above the earth is a function of time given by ( h(t)=-16t^{2}+96t + 4 ). what is the maximum height reached by the ball? hint: find the vertex of the graph of the quadratic function. the maximum height reached by the ball is (square) ft.

Explanation:

Step1: Find the time \(t\) at which the maximum height occurs

For a quadratic function \(y = ax^{2}+bx + c\), the \(x\) - coordinate of the vertex is given by \(t=-\frac{b}{2a}\).
In the function \(h(t)=-16t^{2}+96t + 4\), \(a=-16\) and \(b = 96\).

$$t=-\frac{96}{2\times(-16)}=\frac{-96}{-32}=3$$

Step2: Find the maximum height

Substitute \(t = 3\) into the function \(h(t)\).

$$h(3)=-16\times(3)^{2}+96\times3 + 4$$

$$=-16\times9+288 + 4$$

$$=-144+288+4$$

$$=148$$

Answer:

\(148\)