If a ball is thrown upward at 64 feet per second from the top of a building that is 120 feet high, the height of the ball can be modeled by $\mathrm { S } = 120 + 64 \mathrm { t } - 16 \mathrm { t } ^ { 2 }$ feet, where t is the number of seconds after the ball is thrown. After
How many seconds does the ball reach its maximum height?
A) 2 sec
B) 4 sec
C) 1 sec
D) 3.8 sec

Question

Quizplus · Accepted Answer

The maximum height is reached at the vertex of the parabola represented by the equation. The time at which the maximum height is reached can be found using the formula $t = -\frac{b}{2a}$ for a quadratic equation $ax^2 + bx + c$, where $a = -16$, $b = 64$, and $c = 120$. Plugging these values in gives $t = -\frac{64}{2(-16)} = 2$ seconds.

If a Ball Is Thrown Upward at 64 Feet Per S=120+64t−16t2\mathrm { S } = 120 + 64 \mathrm { t } - 16 \mathrm { t } ^ { 2 }S=120+64t−16t2

If a Ball Is Thrown Upward at 64 Feet Per $\mathrm { S } = 120 + 64 \mathrm { t } - 16 \mathrm { t } ^ { 2 }$