Chapter 14, Problem 71RE

a.

To determine

When does the ball strike the ground?

Answer to Problem 71RE

  $\boxed{7\sec onds}$

Explanation of Solution

Given information:

In physics it is shown that the height $s$ of a ball thrown straight up with an initial speed of $96ft/\sec$ from a rooftop $112feet$ high is

  $s=s(t)=-16t^2+96t+112$

Where $t$ is the elapsed time that the ball is in the air. The ball misses the rooftop on its way down and eventually strikes the ground.

When does the ball strike the ground? That is, how long is the ball in the air?

Calculation:

The height s of a ball thrown up into the air with an initial speed of $96\frac{ft}{\sec }$ from a rooftop $112feet$ high is $s=s(t)=-16t^2+96t+112$ , where $t$ is the elapsed time that the ball is in the air.

The ball eventually strikes the ground on its descent.

The ball strikes the ground when the height of the ball is $0$ . To find out how long the ball is in the air, set the height $s=s(t)=-16t^2+96t+112$ equal to $0$ and solve for $t$

  $s=s(t)=-16t^2+96t+112$

  $\begin{array}{l}0=-16t^2+96t+112\\ 0=-16\left(t^2-6t-7\right)\\ 0=-16\left(t-7\right)\left(t+1\right)\end{array}$

When we set each factor equal to $0$ , we find that $t$ is either $-1or7\sec onds$ . We discard the negative answer because time cannot be negative.

Hence, the ball reaches the ground after $7\sec onds$

b.

To determine

At what time $t$ will the ball pass the rooftop on its way down?

Answer to Problem 71RE

  $\boxed{6\sec onds}$

Explanation of Solution

Given information:

In physics it is shown that the height $s$ of a ball thrown straight up with an initial speed of $96ft/\sec$ from a rooftop $112feet$ high is

  $s=s(t)=-16t^2+96t+112$

Where $t$ is the elapsed time that the ball is in the air. The ball misses the rooftop on its way down and eventually strikes the ground.

At what time $t$ will the ball pass the rooftop on its way down?

Calculation:

The height s of a ball thrown up into the air with an initial speed of $96\frac{ft}{\sec }$ from a rooftop $112feet$ high is $s=s(t)=-16t^2+96t+112$ , where $t$ is the elapsed time that the ball is in the air.

The ball eventually strikes the ground on its descent.

The ball passes the rooftop when it is $112feet$ into the air. Set the formula for the height of the ball equal to $112$ and solve for $t$

  $s(t)=-16t^2+96t+112$

  $112=-16t^2+96t+112$

  $0=-16t^2+96t$

  $0=-16t(t-6)$

When we set each factor equal to $0$ , we find that $t$ is either $0$ or $6$ seconds. We discard the time when $t=0$ because this is when the ball is initially thrown into the air.

Hence, the ball passes the rooftop on its descent towards the ground after $6\sec onds$ .

c.

To determine

What is the average speed of the ball from $t=0$ to $t=2$ ?

Answer to Problem 71RE

  $\boxed{64ft/\sec }$

Explanation of Solution

Given information:

In physics it is shown that the height $s$ of a ball thrown straight up with an initial speed of $96ft/\sec$ from a rooftop $112feet$ high is

  $s=s(t)=-16t^2+96t+112$

Where $t$ is the elapsed time that the ball is in the air. The ball misses the rooftop on its way down and eventually strikes the ground.

What is the average speed of the ball from $t=0$ to $t=2$ ?

Calculation:

The height s of a ball thrown up into the air with an initial speed of $96\frac{ft}{\sec }$ from a rooftop $112feet$ high is $s=s(t)=-16t^2+96t+112$ , where $t$ is the elapsed time that the ball is in the air.

The ball eventually strikes the ground on its descent.

To find the average speed of the ball from $t=0$ to $t=2$ , consider the change in position over the change in time, $\frac{\Delta s}{\Delta t}$ . We need to know the initial position and time as well as the final position and time. The initial time is $t=0$ , the initial position is $112feet$ , and the final time is $t=2$ .

Find the final position by plugging in $t=2$ into $s(t)=-16t^2+96t+112$

  $s(t)=-16t^2+96t+112$

  $\begin{array}{l}s(2)=-16{\left(2\right)}^2+96\left(2\right)+112\\ =-64+192+112\\ =240\end{array}$

Hence, the final position when $t=2$ is $240feet$ . Plug this into the average speed formula to find the average speed of the ball from $t=0$ to $t=2$

  $\begin{array}{l}\frac{\Delta s}{\Delta t}=\frac{s\left(2\right)-s\left(0\right)}{2-0}\\ =\frac{240-112}{2}\end{array}$

  $\begin{array}{l}=\frac{128}{2}\\ =64\end{array}$

Hence, the average speed of ball from $t=0$ to $t=2$ is $64ft/\sec$ .

d.

To determine

What is the instantaneous speed of the ball at time $t$ ?

Answer to Problem 71RE

  $\boxed{(-32t+96)ft/\sec .}$

Explanation of Solution

Given information:

In physics it is shown that the height $s$ of a ball thrown straight up with an initial speed of $96ft/\sec$ from a rooftop $112feet$ high is

  $s=s(t)=-16t^2+96t+112$

Where $t$ is the elapsed time that the ball is in the air. The ball misses the rooftop on its way down and eventually strikes the ground.

What is the instantaneous speed of the ball at time $t$ ?

Calculation:

The height s of a ball thrown up into the air with an initial speed of $96\frac{ft}{\sec }$ from a rooftop $112feet$ high is $s=s(t)=-16t^2+96t+112$ , where $t$ is the elapsed time that the ball is in the air.

The ball eventually strikes the ground on its descent.

To find the instantaneous speed of the ball at time $t$ , let the original position function be

  $s(t_0)=-16t_0{}^2+96t_0+112$ in order to avoid confusion. The instantaneous speed of the ball is the derivative $s\text{'}\left(t\right)={\lim }_{t_0\to t}\frac{s\left(t_0\right)-s\left(t\right)}{t_0-t}$

  ${\lim }_{t_0\to t}\frac{s\left(t_0\right)-s\left(t\right)}{t_0-t}={\lim }_{t_o\to t}\frac{s\left(t_0\right)-s\left(t\right)}{t_0-t}$

  $={\lim }_{t_0\to t}\frac{-16t^2{}_0+96t_0+112-\left(-16t^2+96t+112\right)}{t_0-t}$

  $={\lim }_{t_0\to t}\frac{-16t^2{}_o+96t_0+112+16t^2-96t-112}{t_0-t}$

  $={\lim }_{t_0\to t}\frac{-16(t^2{}_o-t^2-6t_0+6t)}{t_0-t}$

  $={\lim }_{t_0\to t}-16(t_0+t-6)$

  $=-16(t+t-6)$

  $=-16(2t-6)$

  $=-32t+96$

Hence, the instantaneous speed of the ball at time t is $(-32t+96)ft/\sec .$

e.

To determine

What is the instantaneous speed of the ball at $t=2$ ?

Answer to Problem 71RE

  $\boxed{32ft/\sec .}$

Explanation of Solution

Given information:

In physics it is shown that the height $s$ of a ball thrown straight up with an initial speed of $96ft/\sec$ from a rooftop $112feet$ high is

  $s=s(t)=-16t^2+96t+112$

Where $t$ is the elapsed time that the ball is in the air. The ball misses the rooftop on its way down and eventually strikes the ground.

What is the instantaneous speed of the ball at $t=2$ ?

Calculation:

The height s of a ball thrown up into the air with an initial speed of $96\frac{ft}{\sec }$ from a rooftop $112feet$ high is $s=s(t)=-16t^2+96t+112$ , where $t$ is the elapsed time that the ball is in the air.

The ball eventually strikes the ground on its descent.

From part (d) we found that at a general time $t$ , the instantaneous speed $s\text{'}(t)$ was $(-32t+96)ft//\sec$ . To find the instantaneous speed of the ball after $2$ seconds, we can plug in $t=2$ into the equation

  $\begin{array}{l}s\text{'}(t)=-32+96\\ s\text{'}\left(2\right)=-32\left(2\right)+96\end{array}$

  $=32$

Hence, the instantaneous speed of the ball after $2$ seconds is $32ft/\sec .$

f.

To determine

What is the instantaneous speed of the ball equal to zero?

Answer to Problem 71RE

  $\boxed{3\sec onds}$

Explanation of Solution

Given information:

In physics it is shown that the height $s$ of a ball thrown straight up with an initial speed of $96ft/\sec$ from a rooftop $112feet$ high is

  $s=s(t)=-16t^2+96t+112$

Where $t$ is the elapsed time that the ball is in the air. The ball misses the rooftop on its way down and eventually strikes the ground.

What is the instantaneous speed of the ball equal to zero?

Calculation:

The height s of a ball thrown up into the air with an initial speed of $96\frac{ft}{\sec }$ from a rooftop $112feet$ high is $s=s(t)=-16t^2+96t+112$ , where $t$ is the elapsed time that the ball is in the air.

The ball eventually strikes the ground on its descent.

When the instantaneous speed of the ball is $0$ , set the instantaneous speed of the ball equal to $0$ and solve for t

  $\begin{array}{l}s\text{'}(t)=-32+96\\ 0=-32t+96\\ 32t=96\\ t=3\end{array}$

Hence, the ball has an instantaneous speed of $0$ after $3\sec onds$

g.

To determine

What is the instantaneous speed of the ball as it passes the rooftop on the way down?

Answer to Problem 71RE

  $\boxed{-96ft/\sec }$

Explanation of Solution

Given information:

In physics it is shown that the height $s$ of a ball thrown straight up with an initial speed of $96ft/\sec$ from a rooftop $112feet$ high is

  $s=s(t)=-16t^2+96t+112$

Where $t$ is the elapsed time that the ball is in the air. The ball misses the rooftop on its way down and eventually strikes the ground.

What is the instantaneous speed of the ball as it passes the rooftop on the way down?

Calculation:

The height s of a ball thrown up into the air with an initial speed of $96\frac{ft}{\sec }$ from a rooftop $112feet$ high is $s=s(t)=-16t^2+96t+112$ , where $t$ is the elapsed time that the ball is in the air.

The ball eventually strikes the ground on its descent.

From part (b), we found that the ball passes the rooftop on the way down after $6$ seconds. Plug $t=6$ into the instantaneous speed equation $s\text{'}(t)=-32+96$ found in part (d)

  $s\text{'}(t)=-32+96$

  $\begin{array}{l}s\text{'}\left(6\right)=-32\left(6\right)+96\\ =-96\end{array}$

Hence, the instantaneous speed of the ball when it passes the rooftop is $-96ft/\sec$ . The negative sign indicates that the ball is heading downward towards the ground.

h.

To determine

What is the instantaneous speed of the ball when it strikes the ground?

Answer to Problem 71RE

  $\boxed{-128ft/\sec }$

Explanation of Solution

Given information:

In physics it is shown that the height $s$ of a ball thrown straight up with an initial speed of $96ft/\sec$ from a rooftop $112feet$ high is

  $s=s(t)=-16t^2+96t+112$

Where $t$ is the elapsed time that the ball is in the air. The ball misses the rooftop on its way down and eventually strikes the ground.

What is the instantaneous speed of the ball when it strikes the ground?

Calculation:

The height s of a ball thrown up into the air with an initial speed of $96\frac{ft}{\sec }$ from a rooftop $112feet$ high is $s=s(t)=-16t^2+96t+112$ , where $t$ is the elapsed time that the ball is in the air.

The ball eventually strikes the ground on its descent.

From part (a) we found that the ball hits the ground after $7$ seconds. Plug $t=7$ into the instantaneous speed equation $s\text{'}(t)=-32+96$ found in part (d)

  $s\text{'}(t)=-32+96$

  $\begin{array}{l}s\text{'}\left(7\right)=-32\left(7\right)+96\\ =-128\end{array}$

Hence, the instantaneous speed of the ball when it hits the ground is $-128ft/\sec$ .