Question

a ball is thrown vertically upward. after $t$ seconds, its height $h$ (in feet) is given by the function $h(t)=96t - 16t^{2}$. what is the maximum height that the ball will reach?
do not round your answer.
height: $square$ feet

Explanation:

Step1: Identify vertex of quadratic

The function $h(t) = -16t^2 + 96t$ is a downward-opening quadratic ($a<0$), so its vertex gives the maximum. The $t$-coordinate of the vertex of $at^2+bt+c$ is $t=-\frac{b}{2a}$.
$t = -\frac{96}{2(-16)} = \frac{96}{32} = 3$

Step2: Substitute t=3 into h(t)

Calculate height at the vertex time.
$h(3) = 96(3) - 16(3)^2 = 288 - 16(9) = 288 - 144$

Answer:

144 feet