Solve the problem.
-An arrow is fired straight up from the ground with an initial velocity of 240 feet per second. Its height, $\mathrm { s } ( \mathrm { t } )$, in feet at any time $t$ is given by the function $s ( t ) = - 16 t ^ { 2 } + 240 t$. Find the interval of time for which the height of the arrow is greater than 116 feet.
A) between $\frac { 1 } { 2 }$ and $\frac { 29 } { 2 } \mathrm { sec }$
B) after $\frac { 1 } { 2 } \sec$
C) before $\frac { 29 } { 2 } \mathrm { sec }$
D) before $\frac { 1 } { 2 }$ sec or after $\frac { 29 } { 2 } \mathrm { sec }$

Question

Quizplus · Accepted Answer

To find the interval of time for which the height of the arrow is greater than 116 feet, we set the height function greater than 116 and solve for $t$: $-16t^2 + 240t > 116$. Rearranging gives $-16t^2 + 240t - 116 > 0$. Factoring or using the quadratic formula, we find the roots of the equation $-16t^2 + 240t - 116 = 0$ to determine the interval. The correct interval where the height is greater than 116 feet is between $\frac{1}{2}$ and $\frac{29}{2}$ seconds.

Solve the Problem s(t)\mathrm { s } ( \mathrm { t } )s(t)

Solve the Problem $\mathrm { s } ( \mathrm { t } )$