Solve the problem.
-A ball is thrown vertically upward with an initial velocity of 160 feet per second. The distance in feet of the ball from the ground after t seconds is s = 160t - 16t2. For what interval of time is the ball more than 256 above the
Ground?
A) $\{ x \mid 1.5 \mathrm { sec }

Question

Solve the problem.
-A ball is thrown vertically upward with an initial velocity of 160 feet per second. The distance in feet of the ball from the ground after t seconds is s = 160t - 16t2. For what interval of time is the ball more than 256 above the
Ground? 
A) $\{ x \mid 1.5 \mathrm { sec } < \mathrm { x } < 8.5 \mathrm { sec } \}$
B) $\{ x \mid 2 \sec < x < 8 \mathrm { sec } \}$
C) $\{ x \mid 4.5 \mathrm { sec } < x < 5.5 \mathrm { sec } \}$
D) $\{ x \mid 7 \sec < x < 13 \mathrm { sec } \}$

Quizplus · Accepted Answer

To find when the ball is more than 256 feet above the ground, solve the inequality 160t - 16t^2 > 256. This simplifies to -16t^2 + 160t - 256 > 0. Solving the quadratic equation -16t^2 + 160t - 256 = 0 gives the roots t = 2 and t = 8. The ball is above 256 feet between these times, so the interval is 2 < t < 8 seconds.

Solve the Problem {x∣1.5sec<x<8.5sec}\{ x \mid 1.5 \mathrm { sec } < \mathrm { x } < 8.5 \mathrm { sec } \}{x∣1.5sec<x<8.5sec}

Solve the Problem $\{ x \mid 1.5 \mathrm { sec } < \mathrm { x } < 8.5 \mathrm { sec } \}$