Solve the problem.
-A coin is tossed upward with an initial velocity of $16 \mathrm { ft } / \mathrm { sec }$. During what interval of time will the coin be at a height of at least $60 \mathrm { ft }$ ? $\left( \mathrm { h } = - 16 \mathrm { t } ^ { 2 } + \mathrm { v } _ { \mathrm { O } } \mathrm { t } + \mathrm { h } _ { \mathrm { O } } \right.$. $)$
A) $0 \leq t \leq 1$
B) $0 \leq t \leq 3$
C) $2 \leq t \leq 3$
D) $3 \leq t \leq 6$

Question

Quizplus · Accepted Answer

The coin will be at a height of at least 60 ft when $h(t) \ge 60$. So, we solve the inequality $-16t^2 + 16t + 0 \ge 60$. This simplifies to $t^2 - t - 3.75 \le 0$, which factors as $(t - 3)(t + 1.25) \le 0$. So, the coin will be at a height of at least 60 ft when $2 \le t \le 3$.

Solve the Problem 16ft/sec16 \mathrm { ft } / \mathrm { sec }16ft/sec

Solve the Problem $16 \mathrm { ft } / \mathrm { sec }$