Q. 334 - Intermediate Algebra

Question

Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from $48 feet$ above the ground, the function $h\left(t\right)=-16t^2+32t+48$ models the height h, of the ball above the gorund as a function of time t. Find:

Part (a): The zeros of his function which tells us when the ball hits the ground.

Part (b): The time(s) the ball will be $48 feet$ above the ground.

Part (c): The height the ball will be at $t=1$ seconds which is when the ball will be at its highest point.

Step-by-Step Solution

Answer

Part (a): The ball will hit the ground $3$ secs after its thrown.

Part (b): At $0 sec, 2 secs$ time the rocket will be at $48 feet$ above the ground.

Part (c): At $1 sec$, the ball will be at $64 feet$.

Part (a) Step 1. Make the assumption.

Consider the given question,

$h\left(t\right)=-16t^2+32t+48$

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

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

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

$$\begin{gathered} 0=-16t^2+32t+48 \\ 0=-16\left(t^2-2t-3\right) \\ 0=t^2-2t-3 ...... \left(i\right) \end{gathered}$$

Part (a) Step 2. Factor the equation.

On factoring,

$$\begin{gathered} t^2-3t+t-3=0 \\ t\left(t-3\right)+\left(t-3\right)=0 \\ \left(t-3\right)\left(t+1\right)=0 \end{gathered}$$

Use the zero product property, 

$$\begin{gathered} t-3=0 \\ t=3 \end{gathered}$$

Then,

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

As time cannot be negative, so only $t=3$ is considered.

Part (a) Step 3. Check the answers.

Substitute $t=3$ in equation (i),

$$\begin{gathered} 0=3^2-2\left(3\right)-3 \\ 0=9-6-3 \\ 0=0 \end{gathered}$$

This is true.

Hence, the ball will hit the ground $3$ secs after its thrown.

Part (b) Step 1. Make the assumption.

Consider the given question,

$h\left(t\right)=-16t^2+32t+48$

The ball will be $48feet$ above the ground.

Also, $h\left(t\right)=48$

Substitute $h\left(t\right)=48$ in the function,

$$\begin{gathered} 48=-16t^2+32t+48 \\ 0=-16t^2+32t ...... \left(i\right) \\ 0=-16t\left(t-2\right) \end{gathered}$$

Use the zero product property,

Then,

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

Part (b) Step 2. Check the answers.

Substitute $t=2$ in equation (ii),

$$\begin{gathered} 0=-16{\left(2\right)}^2+32\left(2\right) \\ 0=-64+64 \\ 0=0 \end{gathered}$$

This is true.

Substitute $t=0$ in equation (i),

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

This is also true.

Hence, at $0 secs, 2 secs$ time the rocket will be at $48feet$ above the ground.

Part (c) Step 1. Make the assumption.

Consider the given question,

$h\left(t\right)=-16t^2+32t+48$

The height of the ball will be at $t=1$.

Substitute $t=1$ in the function,

$$\begin{gathered} h\left(1\right)=-16{\left(1\right)}^2+32\left(1\right)+48 \\ =-16+32+48 \\ =64 \end{gathered}$$

Part (c) Step 2. Check the answers.

Substitute $h\left(t\right)=64$ in the function,

$$\begin{gathered} 64=-16t^2+32t+48 \\ 0=-16t^2+32t-16 \\ 0=-16\left(t^2-2t+1\right) \\ 0=t^2-2t+1 \\ 0={\left(t-1\right)}^2 \end{gathered}$$

Using the zero product property,

$t=1$

Hence, at $1 sec$, the ball will be at $64 feet$.