Q. 333 - Intermediate Algebra

Question

Juli is going to launch a model rocket in her back yard. When she launches the rocket, the function $h\left(t\right)=-16t^2+32t$ models the height h, of the rocket above the ground as a function of time $t$. Find:

Part (a): The zeros of this function which tells us when the penny will hit the ground.

Part (b): The time the rocket will be $16 feet$ above the ground.

Step-by-Step Solution

Answer

Part (a): The penny will hit the ground $0 secs, 2$ secs after its launched.

Part (b): In $1 sec$ time, the rocket will be at $16 feet$ above the ground.

Part (a) Step 1. Make the assumption.

Consider the given question,

$h\left(t\right)=-16t^2+32t$

To find the zero of the function which tells us when the penny will hit the ground.

Also, $h\left(t\right)=0$.

Substitute $h\left(t\right)=0$ in the function,

$$\begin{gathered} 0=-16t^2+32t ...... \left(i\right) \\ 0=-16t\left(t-2\right) \end{gathered}$$

Use the zero product property,

$$\begin{gathered} -16t=0 \\ t=0 \end{gathered}$$

Then,

$$\begin{gathered} t-2=0 \\ t=2 \end{gathered}$$

Part (a) Step 2. Check the answer.

Substitute the value in equation (i),

$$\begin{gathered} 0=-16{\left(0\right)}^2+32\left(0\right) \\ 0=0 \end{gathered}$$

This is true.

The penny will hit the ground $0 secs, 2 secs$ after its launched.

Part (b) Step 1. Make the assumption.

Consider the given question,

$h\left(t\right)=-16t^2+32t$

The rocket will be $16 feet$ above the ground.

Also, $h\left(t\right)=16$

Substitute $h\left(t\right)=16$ in the function,

$$\begin{gathered} 16=-16t^2+32t \\ 16t^2-32t+16=0 \\ 16\left(t^2-2t+1\right)=0 \\ {t}^2-2t+1=0 ...... \left(i\right) \end{gathered}$$

Part (b) Step 2. Factor the equation.

On factoring,

$$\begin{gathered} t^2-t-t+1=0 \\ t\left(t-1\right)-\left(t-1\right)=0 \\ {\left(t-1\right)}^2=0 \end{gathered}$$

Use the zero product property,

$$\begin{gathered} t-1=0 \\ t=1 \end{gathered}$$

Part (b) Step 3. Check the answers.

Substitute the value in equation (i),

$$\begin{gathered} 1^2-2\left(1\right)+1=0 \\ 1-2+1=0 \\ 0=0 \end{gathered}$$

This is true.

Hence, in $1 sec$ time, the rocket will be at $16 feet$ above the ground.