Q. 462 - Intermediate Algebra

Question

Jing is going to throw a ball from the balcony of her condo. When she throws the ball from $80$ feet above the ground, the function $h\left(t\right)=-16t^2+64t+80$ models the height h, of the ball above the ground as a function of time t. Find:

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

Part (b): The time(s) the ball will be $128$ feet above the ground.

Part (c): The height the ball will be at $t=4secs$.

Step-by-Step Solution

Answer

Part (a): The ball will hit the ground $1$ secs after it is thrown.

Part (b): At $3secs$ time the ball will be at $128$ feet above the ground.

Part (c): At $4secs$ the ball will be at $80$ feet.

Part (a) Step 1. Given information.

Consider the given function,

$h\left(t\right)=-16t^2+64t+80$

To find the zero of the function which tells us when the ball will hit the ground by substituting $h\left(t\right)=0$,

$$\begin{gathered} 0=-16t^2+64t+80 \\ 0=-16\left(t^2-4t-5\right) \\ 0=t^2-4t-5 \end{gathered}$$

Part (a) Step 2. Factor the equation.

On factoring,

$$\begin{gathered} 0=t^2+t-5t-5 \\ 0=t\left(t+1\right)-5\left(t+1\right) \\ 0=\left(t+1\right)\left(t-5\right) \end{gathered}$$

Then,

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

As time cannot be negative. Hence, it is not considered.

Also,

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

Therefore, we can say the ball will hit the ground $1$secs after it is thrown.

Part (b) Step 1. Given information.

To find the ball will be at $128$ feet above the ground by substituting $h\left(t\right)=128$,

$128=-16t^2+64t+80$

Bring all the terms to one side,

$$\begin{gathered} 16t^2-64t+48=0 \\ 16\left(t^2-4t+3\right)=0 \\ {t}^2-4t+3=0 \end{gathered}$$

Part (b) Step 2. Factor the equation.

On factoring,

$$\begin{gathered} t^2-4t+3=0 \\ {t}^2-t-3t+3=0 \\ t\left(t-1\right)-3\left(t-1\right)=0 \\ \left(t-1\right)\left(t-3\right)=0 \end{gathered}$$

Then,

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

Also,

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

Therefore, at $3sec$ time the ball will be at $128$ feet above the ground.

Part (c) Step 1. Given information.

To find the height of the ball at $4secs$, substitute $t=4$ in the function,

$$\begin{gathered} h\left(4\right)=-16{\left(4\right)}^2+64\left(4\right)+80 \\ =-16\left(16\right)+64\left(4\right)+80 \\ =-256+256+80 \\ =80 \end{gathered}$$

Therefore, $4secs$ the ball will be at $80feet$.