the function relating the height of an object off the ground to the time spent falling is a quadratic…

Answer:

C. 50 meters

Explanation:

Step1: Determine the quadratic - function form

The general form of the height - time function for free - fall is $h(t)=h_0 + v_0t-\frac{1}{2}gt^2$, where $h_0$ is the initial height, $v_0$ is the initial velocity, and $g$ is the acceleration due to gravity. Here, $h_0 = 90$ m and $v_0 = 0$ m/s (dropped, not thrown), and $g = 10$ m/s². So $h(t)=90 - 5t^2$.

Step2: Calculate the height at $t = 2$ s

Substitute $t = 2$ into the function $h(t)=90 - 5t^2$. $h(2)=90-5\times2^2$. $h(2)=90 - 5\times4$. $h(2)=90 - 20$. $h(2)=70$ m is wrong. Let's start over.

The general quadratic function for height $h(t)=at^{2}+bt + c$. When $t = 0$, $h(0)=90$, so $c = 90$. When $t=3$, $h(3)=0$. So $0=a\times3^{2}+b\times3 + 90$, which simplifies to $9a + 3b+90 = 0$ or $3a + b=- 30$. Also, since the initial velocity is 0 (dropped), $b = 0$ (the coefficient of the linear term represents the initial velocity in the general height - time quadratic $h(t)=at^{2}+bt + c$). Then $3a=-30$, so $a=- 10$. So $h(t)=-10t^{2}+90$.

Step3: Find the height at $t = 2$ s

Substitute $t = 2$ into $h(t)=-10t^{2}+90$. $h(2)=-10\times2^{2}+90$. $h(2)=-10\times4 + 90$. $h(2)=-40 + 90$. $h(2)=50$ m.