A rock is thrown from one side of a river to another. The function h(t) = -16t^2 + 80t + 30 gives the height in inches of the rock t seconds after it has been thrown.

Answer 1

The function $h(t)=-16t^2+80t+30$ describes the height of the rock over time, where $h(t)$ is the height in inches and $t$ is the time in seconds.

Key Features of the Function:

1. Quadratic Function: This is a quadratic function of the form $h(t)=a{t}^2+bt+c$ where:

   • $a=-16$
   • $b=80$
   • $c=30$

2. Parabola Orientation: Since $a<0$, the parabola opens downwards, indicating that there will be a maximum height achieved before the rock falls back down.

3. Finding the Vertex (Maximum Height): The vertex of a quadratic function given by $h(t)=a{t}^2+bt+c$ occurs at $t=-\frac{b}{2a}$. $$t=-\frac{80}{2\cdot -16}=\frac{80}{32}=2.5\text{ seconds}$$

   To find the maximum height, we plug $t=2.5$ back into the function $h(t)$: $$h(2.5)=-16(2.5{)}^2+80(2.5)+30$$ $$=-16(6.25)+200+30$$ $$=-100+200+30=130\text{ inches}$$

4. Finding when the rock hits the ground: Set $h(t)=0$ to find when the rock hits the ground: $$-16t^2+80t+30=0$$ We can use the quadratic formula $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$:

   • $a=-16$
   • $b=80$
   • $c=30$

   $$t=\frac{-80\pm \sqrt{{80}^2-4\cdot (-16)\cdot 30}}{2\cdot -16}$$ $$t=\frac{-80\pm \sqrt{6400+1920}}{-32}$$ $$t=\frac{-80\pm \sqrt{8320}}{-32}$$ $$t=\frac{-80\pm 91.144}{-32}$$

   Calculating the two possible solutions: $$t_1=\frac{-80+91.144}{-32}=\frac{11.144}{-32}(\text{not physically meaningful})$$ $$t_2=\frac{-80-91.144}{-32}=\frac{-171.144}{-32}\approx 5.34\text{ seconds}$$

Summary:

• Maximum Height: 130 inches at $t=2.5$ seconds.
• Time the rock hits the ground: approximately $t=5.34$ seconds.

If you have specific questions or would like further analysis on this function, feel free to ask!