Solve the problem.
-A coin is tossed upward from a balcony $230 \mathrm { ft }$ high with an initial velocity of $48 \mathrm { ft } / \mathrm { sec }$. During what interval of time will the coin be at a height of at least $70 \mathrm { ft }$ ? $\left( h = - 16 t ^ { 2 } + v _ { 0 } t + h _ { 0 } \right)$
A) $4 \leq t \leq 5$
B) $0 \leq t \leq 5$
C) $5 \leq t \leq 10$
D) $0 \leq t \leq 1$

Question

Quizplus · Accepted Answer

To solve this problem, we use the given formula for the height of the coin at any time $t$, $h = -16t^2 + v_0t + h_0$, where $v_0 = 48 \, \text{ft/sec}$ is the initial velocity and $h_0 = 230 \, \text{ft}$ is the initial height. We are asked to find the interval of time during which the height of the coin is at least $70 \, \text{ft}$. Setting $h \geq 70$, we get:$$ -16t^2 + 48t + 230 \geq 70 $$$$ -16t^2 + 48t + 160 \geq 0 $$$$ -16(t^2 - 3t - 10) \geq 0 $$$$ t^2 - 3t - 10 \leq 0 $$$$ (t - 5)(t + 2) \leq 0 $$Since time cannot be negative, we ignore the solution involving $t + 2$. The inequality is satisfied for $0 \leq t \leq 5$, meaning the coin is at least $70 \, \text{ft}$ high during the interval from $0$ to $5$ seconds after it is tossed.

Solve the Problem 230ft230 \mathrm { ft }230ft High with an Initial Velocity Of

Solve the Problem $230 \mathrm { ft }$ High with an Initial Velocity Of