Solve the problem.
-An object is propelled vertically upward from the top of a 128-foot building. The quadratic function $$s ( t ) = - 16 t ^ { 2 } + 112 t + 128$$ models the ball's height above the ground, s(t), in feet, t seconds after it was thrown. How many seconds does it take until the object finally hits the ground? Round to the nearest tenth of a second if
Necessary.
A) 2 seconds
B) 1.0 seconds
C) 3.5 seconds
D) 8.0 seconds

Question

Quizplus · Accepted Answer

To find when the object hits the ground, set $s(t) = 0$ and solve for $t$. The equation becomes $-16t^2 + 112t + 128 = 0$. Solving this quadratic equation for $t$ gives the time(s) when the object is at ground level. Using the quadratic formula, $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -16$, $b = 112$, and $c = 128$, we find the positive root that makes physical sense for the time. The calculation yields $t \approx 8.0$ seconds, which is when the object hits the ground.

Solve the Problem s(t)=−16t2+112t+128s ( t ) = - 16 t ^ { 2 } + 112 t + 128s(t)=−16t2+112t+128

Solve the Problem $s ( t ) = - 16 t ^ { 2 } + 112 t + 128$