Question

the height of a small rock falling from the top of a 124 - ft - tall building with an initial downward velocity of - 30 ft/sec is modeled by the equation h(t)=-16t² - 30t + 124, where t is the time in seconds. for which interval of time does the rock remain in the air?
o t = 2
o - 2 < t < 0
o 0 < t < 2
o t>2

Explanation:

Step1: Set height equal to 0

Set $h(t)=0$, so we have the quadratic equation $-16t^{2}-30t + 124=0$. Divide through by -2 to simplify: $8t^{2}+15t - 62 = 0$.

Step2: Use quadratic formula

The quadratic formula for $ax^{2}+bx + c=0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 8$, $b=15$, and $c=-62$. First, calculate the discriminant $\Delta=b^{2}-4ac=(15)^{2}-4\times8\times(-62)=225 + 1984=2209$.

Step3: Find the values of t

$t=\frac{-15\pm\sqrt{2209}}{16}=\frac{-15\pm47}{16}$. We get two solutions: $t_1=\frac{-15 + 47}{16}=\frac{32}{16}=2$ and $t_2=\frac{-15 - 47}{16}=\frac{-62}{16}=-\frac{31}{8}$.

Step4: Determine valid time interval

Since time $t\geq0$ in this context, we discard the negative - valued solution. The rock is in the air when $0 < t<2$.

Answer:

C. $0 < t<2$