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 $h(t) = -16t^2 - 30t + 124$, where $t$ is the time in seconds. for which interval of time does the rock remain in the air?

$\bigcirc$ $t = 2$
$\bigcirc$ $-2 < t < 0$
$\bigcirc$ $0 < t < 2$
$\bigcirc$ $t > 2$

Explanation:

Step1: Understand the problem

We need to find when the rock is in the air, which means we need to find the time \( t \) when the height \( h(t)>0 \) (since height can't be negative when it's in the air, and when it hits the ground \( h(t) = 0 \)). The height function is \( h(t)=- 16t^{2}-30t + 124 \). We can find the time when the rock hits the ground by solving \( h(t)=0 \), i.e., \( -16t^{2}-30t + 124 = 0 \). Multiply both sides by - 1 to get \( 16t^{2}+30t - 124=0 \). Divide the entire equation by 2: \( 8t^{2}+15t - 62 = 0 \).

Step2: Solve the quadratic equation

For a quadratic equation \( ax^{2}+bx + c = 0 \), the quadratic formula is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \). Here, \( a = 8 \), \( b = 15 \), \( c=-62 \). First, calculate the discriminant \( D=b^{2}-4ac=(15)^{2}-4\times8\times(-62)=225 + 1984=2209 \). Then \( \sqrt{D}=\sqrt{2209} = 47 \). So \( t=\frac{-15\pm47}{2\times8} \). We have two solutions: \( t_1=\frac{-15 + 47}{16}=\frac{32}{16}=2 \) and \( t_2=\frac{-15 - 47}{16}=\frac{-62}{16}=-3.875 \). Since time \( t\geq0 \) (we can't have negative time in this context), the rock starts at \( t = 0 \) (when it is dropped) and hits the ground at \( t = 2 \) seconds. So the rock is in the air for \( 0

Answer:

\( 0 < t < 2 \) (corresponding to the option "0 < t < 2")