If an object is thrown upward with an initial velocity of $96 \mathrm { ft } / \mathrm { sec }$, its height after $\mathrm { t }$ sec is given by $s ( t ) = 96 \mathrm { t } - 16 \mathrm { t } ^ { 2 }$. Find the maximum height attained by the object. (The object will attain maximum height exactly at the halfway point in terms of the time $t$, where $t = 0$ is at the beginning of the object's flight, and the final time is when the object hits the ground.)
A) 96 ft
B) 128 ft
C) 80 ft
D) 144 ft

Question

Quizplus · Accepted Answer

The maximum height is found by first determining the time at which the object reaches its maximum height. Since the object's height follows a quadratic equation, the maximum height occurs at the vertex of the parabola. Given the formula $s(t) = 96t - 16t^2$, we can find the time to reach maximum height by using the formula $t = \frac{-b}{2a}$, where $a = -16$ and $b = 96$. This gives $t = \frac{-96}{2(-16)} = 3$ seconds. Substituting $t = 3$ back into the equation for $s(t)$ gives the maximum height: $s(3) = 96(3) - 16(3)^2 = 288 - 144 = 144$ ft.

If an Object Is Thrown Upward with an Initial Velocity 96ft/sec96 \mathrm { ft } / \mathrm { sec }96ft/sec

If an Object Is Thrown Upward with an Initial Velocity $96 \mathrm { ft } / \mathrm { sec }$