Chapter 6.3, Problem 40E

Solution

a)

To write: an equation that models the height of the ball as a function of time $t$ .

An equation that models the height of the ball as a function of time $t$ is $y=-16t^2+80t+5$ .

Given information:

The initial velocity is $80$ ft/sec. The height is $5$ ft.

Formula used:

The formula to find the height of the ball:

  $y=-16t^2+v_{0}t+h$

Calculation:

Substitute $5$ for $h$ and $80$ in $v_0$ in $y=-16t^2+v_{0}t+h$ .

  $y=-16t^2+80t+5$

Hence, an equation that models the height of the ball as a function of time $t$ is $y=-16t^2+80t+5$ .

b)

To simulate: the pop-up.

The graph shown below depicts how the ball begins at a height of $5$ ft, rises to over $100$ ft.

  

The graph below shows the fall of the ball back to the ground.

  

Given information:

The initial velocity is $80$ ft/sec. The height is $5$ ft.

Calculation:

The graph shown below depicts how the ball begins at a height of $5$ ft, rises to over $100$ ft.

  

The graph below shows the fall of the ball back to the ground.

  

c)

To graph: height against time.

The graph is shown below.

  

Given information:

The initial velocity is $80$ ft/sec. The height is $5$ ft.

Calculation:

The graph is shown below.

  

d)

To find: the height of the ball after $4$ seconds.

After $4$ seconds, the height of the ball was $69$ ft.

Given information:

The initial velocity is $80$ ft/sec. The height is $5$ ft.

Formula used:

The formula to find the height of the ball:

  $y=-16t^2+v_{0}t+h$

Calculation:

Substitute $80$ for $v_0$ , $4$ for $h$ and $4$ for $t$ in $y=-16t^2+v_{0}t+h$

  $\begin{array}{c}y=-16{\left(4\right)}^2+80\left(4\right)+5\\ =-16\left(16\right)+320+5\\ =-256+320+5\\ =69\end{array}$

After $4$ seconds, the height of the ball was $69$ ft.

e)

To find: the maximum height of the ball and how many seconds does it take to reach its maximum height.

The time taken to reach the maximum height $105$ feet is $2.5$ seconds.

Given information:

The initial velocity is $80$ ft/sec. The height is $5$ ft.

Formula used:

The formula to find the height of the ball:

  $y=-16t^2+v_{0}t+h$

Calculation:

  

From the graph it is evident that $t=2.5$ and the maximum height of the ball is $105$ feet.

Therefore, the time taken to reach the maximum height $105$ feet is $2.5$ seconds.