Solve the problem.
-A coin is tossed upward from a balcony $390 \mathrm { ft }$ high 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 $70 \mathrm { ft }$ ?
$\left( h = - 16 t ^ { 2 } + v _ { O } t + h _ { O } \cdot \right)$
A) $0 \leq t \leq 1$
B) $0 \leq t \leq 5$
C) $4 \leq \mathrm { t } \leq 5$
D) $5 \leq t \leq 10$

Question

Quizplus · Accepted Answer

Given the equation for height $h = -16t^2 + 16t + 390$, we set $h = 70$ to find the time interval when the coin is at least 70 ft high. Solving $70 = -16t^2 + 16t + 390$ gives two time values that frame the interval where the height is above 70 ft. The correct interval from solving this equation is $0 \leq t \leq 5$, indicating the coin is at least 70 ft high between 0 and 5 seconds after being tossed.

Solve the Problem 390ft390 \mathrm { ft }390ft High with an Initial Velocity Of

Solve the Problem $390 \mathrm { ft }$ High with an Initial Velocity Of