The equation below is in the form $h = a t ^ { 2 } + b t + c$ The input is the time t, in seconds. The output is the height h of an object tossed straight up from a height c, in feet, with an initial velocity b, in feet per second. $- 16 t ^ { 2 } + 15 t + 110 = 0$ Solve the equation for t, using the quadratic formula. Round to the nearest hundredth. Which answers are acceptable in the setting?
A) $t = 2.02 \mathrm { sec } \text {, discard } t = - 3.13$
B) $t = 3.13 \mathrm { sec } , \text { discard } t = - 2.19$
C) $t = 1.78 \mathrm { sec } , \text { discard } t = - 2.02$
D) $t = 3.13 \mathrm { sec } , \text { discard } t = - 2.02$
E) no real-number solution

Question

Quizplus · Accepted Answer

Using the quadratic formula, $$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ where $a=-16$, $b=15$, and $c=110$. Plugging in these values, we get $$t=\frac{-15\pm\sqrt{15^2-4(-16)(110)}}{2(-16)}.$$ Simplifying, we get $$t=\frac{15\pm\sqrt{3045}}{-32}.$$ Rounding to the nearest hundredth, we get $t=-2.19$ or $t=3.13$. Discarding the negative solutions, we are left with only one acceptable answer, which is choice B, $t=3.13	ext{ sec}$.

The Equation Below Is in the Form h=at2+bt+ch = a t ^ { 2 } + b t + ch=at2+bt+c

The Equation Below Is in the Form $h = a t ^ { 2 } + b t + c$