Solve the problem
-A person standing close to the edge on top of a 128-foot building throws a baseball vertically upward. The quadratic function $s ( t ) = - 16 t ^ { 2 } + 64 t + 128$ models the ball's height above the ground, $s ( t )$, in feet, $t$ seconds after it was thrown. After how many seconds does the ball reach its maximum height? Round to the nearest tenth of a second if necessary.
A) $1.5$ seconds
B) 192 seconds
C) 2 seconds
D) $5.5$ seconds

Question

Quizplus · Accepted Answer

The maximum height of a projectile in a quadratic equation of the form $s(t) = -at^2 + bt + c$ is reached at $t = \frac{-b}{2a}$. Here, $a = 16$ and $b = 64$, so $t = \frac{-64}{2(-16)} = 2$ seconds.

Solve the Problem -A Person Standing Close to the Edge s(t)=−16t2+64t+128s ( t ) = - 16 t ^ { 2 } + 64 t + 128s(t)=−16t2+64t+128

Solve the Problem
-A Person Standing Close to the Edge $s ( t ) = - 16 t ^ { 2 } + 64 t + 128$