Chapter 2.1, Problem 61E

Solution

a.

To Determine:

The maximum height of the baseball.

The maximum height of the baseball is approximately $215.25\text{ ft}$ .

Given:

Initial height of baseball $s_0=83\text{ ft}$ .

Initial vertical velocity of baseball $v_0=92\text{ ft/sec}$ .

Concepts Used:

The height $s$ and vertical velocity $v$ of an object in free fall are modeled by the equations $s\left(t\right)=-\frac{1}{2}g{t}^2+v_{0}t+s_0$ and $v\left(t\right)=-gt+v_0$ , where $t$ is time (in seconds), $g\approx 32{\text{ ft/sec}}^2\approx 9.8{\text{ m/sec}}^2$ is the acceleration due to gravity , $v_0$ is the initial vertical velocity of the object, and $s_0$ is its initial height.

Plotting a function on graph and reading its extrema.

Calculations:

Substitute the values of initial height of baseball $s_0=83\text{ ft}$ , initial vertical velocity of baseball $v_0=92\text{ ft/sec}$ , and the acceleration due to gravity $g\approx 32{\text{ ft/sec}}^2$ in the equations $s\left(t\right)=-\frac{1}{2}g{t}^2+v_{0}t+s_0$ and $v\left(t\right)=-gt+v_0$ to get the models for the present situation:

  $\begin{array}{l}s\left(t\right)=-\frac{1}{2}\left(32\right)t^2+\left(92\right)t+83\\ s\left(t\right)=-16t^2+92t+83\end{array}$

and

  $\begin{array}{l}v\left(t\right)=-gt+v_0\\ v\left(t\right)=-32t+92\end{array}$

Note that it is implicit that distances are measured in $\text{ft}$ and time is measured in $\text{sec}$ .

Plot the function $s\left(t\right)=-16t^2+92t+83$ that models the height. Indicate the point of maximum height.

  

The function $s\left(t\right)=-16t^2+92t+83$ is plotted above. The graph shows a maximum for $s\left(t\right)=215.25$ when $t=2.88$ .

Conclusion:

The maximum height of the baseball is approximately $215.25\text{ ft}$ .

b.

To Determine:

The duration for which the ball is in the air.

The ball stays in the air for about $6.54$ seconds.

Given:

Initial height of baseball $s_0=83\text{ ft}$ .

Initial vertical velocity of baseball $v_0=92\text{ ft/sec}$ .

Known from previous part:

The function $s\left(t\right)=-16t^2+92t+83$ models the height of the ball.

Concepts Used:

The duration an object free falling under gravity stays in air is the positive time for which the height becomes zero (time elapsed before it touches ground).

Plotting a function and determining its roots.

Calculations:

Plot the function $s\left(t\right)=-16t^2+92t+83$ that models the height. Indicate the point where the height becomes zero for a positive value of $t$ .

  

The function $s\left(t\right)=-16t^2+92t+83$ is plotted above. The graph shows that the curve intersects the $t$ axis at $\left(6.54,0\right)$ , that is, when $t=6.54$ .

Conclusion:

The ball stays in the air for about $6.54$ seconds.

c.

To Calculate:

The vertical velocity of the ball when it hits ground.

The vertical velocity is $\text{117}\text{.28 ft/sec}$ in the downward direction when the ball hits the ground.

Given:

Initial height of baseball $s_0=83\text{ ft}$ .

Initial vertical velocity of baseball $v_0=92\text{ ft/sec}$ .

Known from previous part:

The function $v\left(t\right)=-32t+92$ models the vertical velocity of the ball.

The ball hits the ground when $t=6.54$ .

Concepts Used:

Substitution.

Calculations:

Substitute $t=6.54$ in the equation $v\left(t\right)=-32t+92$ that models the vertical velocity of the ball.

  $\begin{array}{c}v\left(t\right)=-32t+92\\ v\left(6.54\right)=-32\times 6.54+92\\ =-209.28+92\\ =-117.28\end{array}$

Thus, $v\left(6.54\right)=-117.28$ .

Conclusion:

The vertical velocity is $\text{117}\text{.28 ft/sec}$ in the downward direction when the ball hits the ground.