Question

1. a rock is thrown directly upward from a cliff above a beach. the height of the rock (in feet) above the beach (h(t)) after t seconds is given by the equation: h(t)=-16t² + 96t + 200

a) how high does the rock go? how long does it take to get there?
b) at what height was the rock thrown from?
c) how long did it take the rock to hit the beach?
d) when was the rock 300 feet above the beach?

Explanation:

Step1: Find the vertex for part a

The height - function is a quadratic function $h(t)=-16t^{2}+96t + 200$, where $a=-16$, $b = 96$, $c = 200$. The time $t$ at the vertex of a quadratic function $y = ax^{2}+bx + c$ is given by $t=-\frac{b}{2a}$.
$t=-\frac{96}{2\times(-16)}=\frac{-96}{-32}=3$ seconds.

Step2: Find the maximum height for part a

Substitute $t = 3$ into the height - function $h(t)=-16t^{2}+96t + 200$.
$h(3)=-16\times3^{2}+96\times3 + 200=-16\times9 + 288+200=-144 + 288+200=344$ feet.

Step3: Find the initial height for part b

The initial height is when $t = 0$. Substitute $t = 0$ into $h(t)=-16t^{2}+96t + 200$.
$h(0)=-16\times0^{2}+96\times0 + 200 = 200$ feet.

Step4: Find the time when the rock hits the beach for part c

Set $h(t)=0$, so $-16t^{2}+96t + 200 = 0$. Divide through by - 8 to simplify: $2t^{2}-12t - 25 = 0$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 2$, $b=-12$, $c=-25$.
$t=\frac{12\pm\sqrt{(-12)^{2}-4\times2\times(-25)}}{2\times2}=\frac{12\pm\sqrt{144 + 200}}{4}=\frac{12\pm\sqrt{344}}{4}=\frac{12\pm2\sqrt{86}}{4}=\frac{6\pm\sqrt{86}}{2}$.
We take the positive root $t=\frac{6+\sqrt{86}}{2}\approx7.636$ seconds.

Step5: Find the time when the rock is 300 feet above the beach for part d

Set $h(t)=300$, so $-16t^{2}+96t + 200 = 300$.
Rearrange to get $-16t^{2}+96t - 100 = 0$. Divide through by - 4: $4t^{2}-24t + 25 = 0$.
Using the quadratic formula $t=\frac{24\pm\sqrt{(-24)^{2}-4\times4\times25}}{2\times4}=\frac{24\pm\sqrt{576 - 400}}{8}=\frac{24\pm\sqrt{176}}{8}=\frac{24\pm4\sqrt{11}}{8}=\frac{6\pm\sqrt{11}}{2}$.
$t_1=\frac{6+\sqrt{11}}{2}\approx4.658$ seconds and $t_2=\frac{6 - \sqrt{11}}{2}\approx1.342$ seconds.

Answer:

a) The rock goes 344 feet high and it takes 3 seconds to get there.
b) The rock was thrown from a height of 200 feet.
c) It takes approximately 7.636 seconds for the rock to hit the beach.
d) The rock is 300 feet above the beach at approximately $t_1=\frac{6+\sqrt{11}}{2}\approx4.658$ seconds and $t_2=\frac{6 - \sqrt{11}}{2}\approx1.342$ seconds.