A coin is tossed upward from a balcony 200 feet high with an initial velocity of 48 feet per second, then its height after t seconds is given by the equation $h ( t ) = - 16 t ^ { 2 } + 48 t + 200$
During what interval of time will the coin be at a height of at least 40 ft?
A) $0 \leq t \leq 5$
B) $4 \leq t \leq 5$
C) $0 \leq t \leq 1$
D) $5 \leq t \leq 10$

Question

A coin is tossed upward from a balcony 200 feet high with an initial velocity of 48 feet per second, then its height after  t  seconds is given by the equation $h ( t ) = - 16 t ^ { 2 } + 48 t + 200$
During what interval of time will the coin be at a height of at least  40 ft?
A) $0 \leq t \leq 5$
B) $4 \leq t \leq 5$
C) $0 \leq t \leq 1$
D) $5 \leq t \leq 10$

Quizplus · Accepted Answer

To find the interval of time during which the coin will be at a height of at least 40 ft, we need to solve the inequality $-16t^2 + 48t + 200 \geq 40$. Subtracting 40 from both sides gives $-16t^2 + 48t + 160 \geq 0$. Factoring out -16 gives $16t^2 - 48t - 160 \leq 0$, which simplifies to $t^2 - 3t - 10 \leq 0$. Factoring the quadratic equation gives $(t-5)(t+2) \leq 0$. Since time cannot be negative, we only consider $t \geq 0$, which gives the solution $0 \leq t \leq 5$.

A Coin Is Tossed Upward from a Balcony 200 Feet h(t)=−16t2+48t+200h ( t ) = - 16 t ^ { 2 } + 48 t + 200h(t)=−16t2+48t+200

A Coin Is Tossed Upward from a Balcony 200 Feet $h ( t ) = - 16 t ^ { 2 } + 48 t + 200$