Chapter 8.2, Problem 33E

a.

To determine

To write: the function $T\left(x\right)=h\left(x\right)-g\left(x\right)$ and what does it represent.

Answer to Problem 33E

  $T\left(x\right)=-44$

This function represents that the distance between the two balls at any point while falling is always 44 feet. And the negative sign here is indicating that your friend’s ball is above your ball.

Explanation of Solution

Given:

The height (in feet) of your ball x seconds after it is dropped is given by the function:

  $h\left(x\right)=-16x^2+256$......(1)

The height (in feet) of your friend’s ball x seconds after it is dropped is given by the function:

  $g\left(x\right)=-16x^2+300$......(2)

Calculation:

Consider,

  $\begin{array}{c}T\left(x\right)=h\left(x\right)-g\left(x\right)\\ =-16x^2+256-\left(-16x^2+300\right)\\ =-16x^2+256+16x^2-300\\ =-44\end{array}$

Thus,

  $T\left(x\right)=-44$

This function represents that the distance between the two balls at any point while falling is always 44 feet. And the negative sign here is indicating that your friend’s ball is above your ball.

  

b.

To determine

the height of your friend’s ball when your ball hits the ground.

Answer to Problem 33E

Your friend’s ball is at a height of 44 feet when your ball hits the ground.

Explanation of Solution

Given:

The height (in feet) of your ball x seconds after it is dropped is given by the function:

  $h\left(x\right)=-16x^2+256$......(1)

The height (in feet) of your friend’s ball x seconds after it is dropped is given by the function:

  $g\left(x\right)=-16x^2+300$......(2)

Calculation:

The time when your ball hits the ground is found by substituting $h\left(x\right)=0$ in equation (1),

  $\begin{array}{c}h\left(x\right)=-16x^2+256\\ 0=-16x^2+256\\ 16x^2=256\end{array}$

  $\begin{array}{c}{x}^2=\frac{256}{16}\\ x^2=16\\ x=\pm 4\end{array}$

Now, here x represents time and time can never be negative, so solution is $x=4$ .

Now the height of friend’s ball at time when $x=4$ is found by substituting $x=4$ in equation (2),

  $\begin{array}{c}g\left(x\right)=-16{\left(4\right)}^2+300\\ =-256+300\\ =44\end{array}$

So, your friend’s ball is at a height of 44 feet when your ball hits the ground.

This can also be verified by observing the graphs of the given function as shown below: