A coin is tossed upward from a balcony 390 ft high (h₀) with an initial velocity (v₀) of 16 ft/sec , according to the formula $h ( t ) = - 16 t ^ { 2 } + v _ { 0 } t + h 0$. where t is time in seconds. During what interval of time will the coin be at a height of at least 70 ft ? A) $4 \sec \leq t \leq 5 \mathrm { sec }$ B) $0 \mathrm { sec } \leq \mathrm { t } \leq 5 \mathrm { sec }$ C) $5 \mathrm { sec } \leq \mathrm { t } \leq 10 \mathrm { sec }$ D) $0 \sec \leq t \leq 1 \mathrm { sec }$

Question

Quizplus · Accepted Answer

To find the interval of time when the coin is at least 70 ft high, we substitute $h(t) = 70$ into the given formula and solve for $t$. The equation becomes $70 = -16t^2 + 16t + 390$. Rearranging gives $16t^2 - 16t - 320 = 0$, which simplifies to $t^2 - t - 20 = 0$. Factoring or using the quadratic formula yields two time values, which represent the times when the coin is exactly 70 ft high during its ascent and descent. The interval between these times, including the initial toss and the moment it reaches 70 ft on its way down, covers from $0 \sec$ to approximately $5 \sec$, indicating the coin is at or above 70 ft within this time frame.

A Coin Is Tossed Upward from a Balcony 390 h(t)=−16t2+v0t+h0h ( t ) = - 16 t ^ { 2 } + v _ { 0 } t + h 0h(t)=−16t2+v0​t+h0

A Coin Is Tossed Upward from a Balcony 390 $h ( t ) = - 16 t ^ { 2 } + v _ { 0 } t + h 0$