If a ball is thrown upward at 64 feet per second from the top of a building that is 110 feet high, the height of the ball can be modeled by $S = 110 + 64 t - 16 t ^ { 2 }$ feet, where t is the number of seconds after the ball is thrown. What
Is the ball's maximum height?
A) 302 ft
B) 110 ft
C) 158 ft
D) 174 ft

Question

Quizplus · Accepted Answer

The maximum height of the ball can be found by determining the vertex of the parabola represented by the equation $S = 110 + 64t - 16t^2$. The t-coordinate of the vertex is given by $-\frac{b}{2a}$, where $a = -16$ and $b = 64$. Thus, $t = -\frac{64}{2(-16)} = 2$. Substituting $t = 2$ back into the equation gives $S = 110 + 64(2) - 16(2)^2 = 110 + 128 - 64 = 174$ feet.

If a Ball Is Thrown Upward at 64 Feet Per S=110+64t−16t2S = 110 + 64 t - 16 t ^ { 2 }S=110+64t−16t2

If a Ball Is Thrown Upward at 64 Feet Per $S = 110 + 64 t - 16 t ^ { 2 }$