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 $70 \mathrm { ft }$ ?
$\left( h = - 16 t ^ { 2 } + v _ { O ^ { - } } t + h _ { O ^ { - } } \right)$
A) $0 \leq t \leq 1$
B) $0 \leq t \leq 5$
C) $4 \leq t \leq 5$
D) $5 \leq t \leq 10$

Question

Quizplus · Accepted Answer

To find the interval of time during which the coin is at a height of at least 70 ft, we use the given equation for height $h = -16t^2 + 16t + 390$, where $h$ is the height in feet, $t$ is the time in seconds, $v_0 = 16 \mathrm{ft/s}$ is the initial velocity, and $h_0 = 390 \mathrm{ft}$ is the initial height. We set $h \geq 70$ and solve for $t$. The quadratic equation represents a parabola opening downwards, so the coin starts at 390 ft, goes up, and then comes down. The time interval when the height is at least 70 ft includes the ascent and descent until it reaches 70 ft again. Solving the inequality $ -16t^2 + 16t + 390 \geq 70 $ gives the interval where this condition is true, which is from the start until just before it hits the ground, thus covering the interval from $0$ to approximately $5$ seconds, considering the time it takes for the coin to first ascend slightly and then descend past 70 ft on its way down.

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