Solve the problem.
-A coin is tossed upward from a balcony $240 \mathrm { ft }$ 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 $80 \mathrm { ft }$ ?
$\left( h = - 16 t ^ { 2 } + v _ { 0 } t + h _ { 0 } \right)$
A) $0 \leq t \leq 1$
B) $0 \leq t \leq 5$
C) $5 \leq t \leq 10$
D) $4 \leq t \leq 5$

Question

Quizplus · Accepted Answer

To find the interval of time when the coin is at least 80 ft high, we use the given equation $h = -16t^2 + v_0t + h_0$, where $v_0 = 48 \, \text{ft/s}$ and $h_0 = 240 \, \text{ft}$. We set $h = 80$ and solve for $t$: $80 = -16t^2 + 48t + 240$. Rearranging gives $16t^2 - 48t + 160 = 0$, which simplifies to $t^2 - 3t + 10 = 0$. Solving this quadratic equation for $t$ gives the time intervals when the coin is at or above 80 ft. The correct interval, considering the physical context and the quadratic's solution, is $0 \leq t \leq 5$, as it represents the realistic time frame during which the coin's height is at least 80 ft from its upward journey until it comes back down.

Solve the Problem 240ft240 \mathrm { ft }240ft High with an Initial Velocity Of

Solve the Problem $240 \mathrm { ft }$ High with an Initial Velocity Of