Q 6.109 - Intermediate Algebra

Question

Genevieve is going to throw a rock from the top a trail overlooking the ocean. When she throws the rock upward from $160$ feet above the ocean, the function $h\left(t\right)=-16t^2+48t+160$ models the height, h, of the rock above the ocean as a function of time, t. Find:

(a) the zeros of this function which tell us when the rock will hit the ocean

(b) when the rock will be $160$ feet above the ocean.

(c) the height of the rock at $t=1.5$ seconds.

Step-by-Step Solution

Answer

Part a. The rock will be hit the ocean $5$seconds after it is thrown.

Part b. The rock will be $160$feet above the ocean after $3 seconds.$

Part c. The height of the rock at $t=1.5$ seconds is $196 feet.$

Part a Step 1. Given Information

The given function is

$h\left(t\right)=-16t^2+48t+160$

We have to find the zeros of this function which tell us when the rock will hit the ocean.

Part a Step 2. Finding the zeros of the function

To find the zeros of the function,

Put $h\left(t\right)=0$

So, $$\begin{gathered} -16t^2+48t+160=0 \\ -16(t^2-3t-10)=0 \\ -16\left[\right(t^2-5t+2t-10\left)\right]=0 \\ -16\left[t\right(t-5)+2(t-5\left)\right]=0 \\ -16\left[\right(t+2\left)\right(t-5\left)\right]=0 \end{gathered}$$

Part a Step 3. Use the Zero Product Property

The equation we get is

$-16\left[\right(t+2\left)\right(t-5\left)\right]=0$

By using the zero product property

$-16\ne 0$ , $$\begin{gathered} t+2=0 \\ t=-2 \end{gathered}$$ , $$\begin{gathered} t-5=0 \\ t=5 \end{gathered}$$

The $t=5$ tells us the rock will hit the ocean $5$ seconds after it is thrown. Since time cannot be negative, the result $t=-2$ is discarded.

Part b Step 1. Given Information

Genevieve throws the rock upward

from $160$ feet above the ocean, the function $h\left(t\right)=-16t^2+48t+160$ 

models the height, h, of the rock above

the ocean as a function of time, t.

We have to find when the rock will be $160$feet above the ocean.

Part b Step 2. To find when the rock will be 160 feet above the ocean

Put $h\left(t\right)=160$

Now, substitute the function

$$\begin{gathered} -16t^2+48t+160=160 \\ -16t^2+48t=0 \\ 48t=16t^2 \\ 3=t \end{gathered}$$

When Genevieve throws the rock, the rock will be $160$ feet above the ocean after $3 seconds$.

Part c Step 1. Given Information

Genevieve throws the rock upward

from $160$ feet above the ocean, the function  $h\left(t\right) = -16t^2 + 48t + 160$

models the height, h, of the rock above

the ocean as a function of time, t.

We have to find the height of the rock at $t=1.5$ seconds.

Part c Step 2. To find the height of the rock at t = 1 . 5 seconds

Substitute $1.5$ for t in h(t)

Therefore, $$\begin{gathered} h(1.5)=-16{(1.5)}^2+ 48(1.5) +160 \\ h(1.5)=-16(2.25)+ 72 + 160 \\ h(1.5) = -36 + 232 \\ h(1.5) =196 \end{gathered}$$

After $1.5$ seconds, the rock will be at $196$ feet.