Question

if a projectile is fired straight upward from the ground with an initial speed of 128 feet per second, then its height h in feet after t seconds is given by the function h(t) = -16t² + 128t. find the maximum height of the projectile. the maximum height of the projectile is (simplify your answer.)

Explanation:

Step1: Find the vertex of the quadratic function

The height function is \(h(t)=-16t^{2}+128t\), which is a quadratic function in the form \(y = ax^{2}+bx + c\) (\(a=-16\), \(b = 128\), \(c = 0\)). The \(t\) - coordinate of the vertex of a quadratic function \(y=ax^{2}+bx + c\) is given by \(t=-\frac{b}{2a}\).

$$t=-\frac{128}{2\times(-16)}=\frac{- 128}{-32}=4$$

Step2: Find the maximum height

Substitute \(t = 4\) into the height function \(h(t)\).

$$h(4)=-16\times4^{2}+128\times4=-16\times16 + 512=-256+512 = 256$$

Answer:

\(256\)