Solve the problem.
-A professional fireworks team shoots an 8-in. mortar straight upwards from ground level with an initial speed of $256 \mathrm { ft } / \mathrm { sec }$.
a. Write a function modeling the vertical position $s ( t )$ (in $\mathrm { ft }$ ) of the shell at a time $t$ seconds after launch.
b. The spectators can see the shell rising once it clears a 768-ft tree line. From what period of time after launch is the shell visible?
A) a. $s ( t ) = - 16 t ^ { 2 } + 256 t$
b. The spectators can see the shell between 4 and $12 \mathrm { sec }$ after launch.
B) a. $s ( t ) = - 16 t ^ { 2 } + 256 t + 768$
b. The spectators can see the shell between 4 and $24 \mathrm { sec }$ after launch.
C) a. $s ( t ) = - 32 t ^ { 2 } + 256 t$
b. The spectators can see the shell between 4 and $12 \mathrm { sec }$ after launch.
D) a. $s ( t ) = - 32 t ^ { 2 } + 256 t$
b. The spectators can see the shell between 4 and 24 sec after launch.

Question

Quizplus · Accepted Answer

a. The function $s(t) = -16t^2 + 256t$ models the vertical position of the shell using the formula $s(t) = -16t^2 + vt$, where $v$ is the initial velocity (256 ft/sec) and the acceleration due to gravity is $32 ft/sec^2$ (the negative sign indicates a downward acceleration, hence the $16$ in the formula as $1/2$ of $32$). b. To find when the shell is above 768 ft, set $s(t) = 768$ and solve for $t$, which gives the time interval when the shell is visible above the tree line.

Solve the Problem 256ft/sec256 \mathrm { ft } / \mathrm { sec }256ft/sec

Solve the Problem $256 \mathrm { ft } / \mathrm { sec }$