Solve the problem. -A ball is thrown vertically upward with an initial velocity of 128 feet per second. The distance in feet of the ball from the ground after t seconds is $$\mathrm { s } = 128 \mathrm { t } - 16 \mathrm { t } ^ { 2 }$$ . For what intervals of time is the ball less than 112 above the Ground (after it is tossed until it returns to the ground)? A) $\{ x \mid 0 \mathrm { sec } 8 \mathrm { sec } \}$ B) $\{ \mathrm { x } \mid 0 \mathrm { sec } 8 \mathrm { sec } \}$ C) $\{ x \mid 0 \sec 8 \mathrm { sec } \}$ D) $\{ \mathrm { x } \mid 1 \mathrm { sec }

Question

Solve the problem. -A ball is thrown vertically upward with an initial velocity of 128 feet per second. The distance in feet of the ball from the ground after t seconds is $$\mathrm { s } = 128 \mathrm { t } - 16 \mathrm { t } ^ { 2 }$$ . For what intervals of time is the ball less than 112 above the Ground (after it is tossed until it returns to the ground)? A) $\{ x \mid 0 \mathrm { sec } < x < 3.5 \mathrm { sec }$ and $4.5 \mathrm { sec } > x > 8 \mathrm { sec } \}$ B) $\{ \mathrm { x } \mid 0 \mathrm { sec } < \mathrm { x } < 0.5 \mathrm { sec }$ and $7.5 \mathrm { sec } > x > 8 \mathrm { sec } \}$ C) $\{ x \mid 0 \sec < x < 1 \mathrm { sec }$ and $7 \sec > x > 8 \mathrm { sec } \}$ D) $\{ \mathrm { x } \mid 1 \mathrm { sec } < \mathrm { x } < 7 \mathrm { sec } \}$

Quizplus · Accepted Answer

The ball's height is given by $ s = 128t - 16t^2 $. Setting $ s < 112 $, we solve $ 128t - 16t^2 < 112 $ to find the intervals where the ball is less than 112 feet above the ground. Solving the inequality gives $ 0 < t < 1 $ and $ 7 < t < 8 $.

Solve the Problem s=128t−16t2\mathrm { s } = 128 \mathrm { t } - 16 \mathrm { t } ^ { 2 }s=128t−16t2

Solve the Problem $\mathrm { s } = 128 \mathrm { t } - 16 \mathrm { t } ^ { 2 }$