Question

if a football is kicked straight up with an initial velocity of 128 ft/sec from a height of 5 ft, then its height above the earth is a function of time given by h(t)= - 16t^2 + 128t + 5. what is the maximum height reached by this ball? hint: find the vertex of the graph of the quadratic function.
the maximum height reached by the ball is ft.

Explanation:

Step1: Identify coefficients

For the quadratic function $h(t)=-16t^{2}+128t + 5$, $a=-16$, $b = 128$, $c = 5$.

Step2: Find the time $t$ at vertex

The formula for the $t$-coordinate of the vertex of a quadratic function $y = ax^{2}+bx + c$ is $t=-\frac{b}{2a}$. So $t=-\frac{128}{2\times(-16)}=\frac{-128}{-32}=4$.

Step3: Find the maximum height

Substitute $t = 4$ into the function $h(t)$. $h(4)=-16\times4^{2}+128\times4 + 5=-16\times16+512 + 5=-256+512+5=261$.

Answer:

261