Solve the problem.
-A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance in feet of the ball from the ground after $t$ seconds is $\mathrm { s } = 96 \mathrm { t } - 16 \mathrm { t } ^ { 2 }$. For what interval of time is ball more than 128 above the ground?
A) between 2 and 4 seconds
B) between $1.5$ and $4.5$ seconds
C) between 5 and 7 seconds
D) between $2.5$ and $3.5$ seconds

Question

Quizplus · Accepted Answer

To find when the ball is more than 128 feet above the ground, we set $s > 128$: $96t - 16t^2 > 128$. Rearranging gives $16t^2 - 96t + 128 < 0$, which simplifies to $t^2 - 6t + 8 < 0$. Factoring yields $(t-2)(t-4) < 0$, indicating the ball is more than 128 feet above the ground between $t = 2$ and $t = 4$ seconds.

Solve the Problem ttt Seconds Is s=96t−16t2\mathrm { s } = 96 \mathrm { t } - 16 \mathrm { t } ^ { 2 }s=96t−16t2

Solve the Problem $t$ Seconds Is $\mathrm { s } = 96 \mathrm { t } - 16 \mathrm { t } ^ { 2 }$