Chapter 11.3, Problem 71E

(a).

To determine

To calculate: The formula for the instantaneous rate of change of the balloon.

Answer to Problem 71E

The formula of instantaneous rate of change of the balloon is $s`\left(t\right)=64-32t$

Explanation of Solution

Given information: The displacement s, $s\left(t\right)=-16t^2+64t+80$

Formula used: $\frac{d}{dx}\left(x^n\right)=n{x}^{n+1}$

Differentiate the equation of displacement ( s ) with respect to $\text{'}t\text{'}$

Calculation:

  $\begin{array}{l}\frac{d}{dt}s=-16\left(2t^{2-1}\right)+64\left(t^{1-1}\right)+0\\ \Rightarrow s`\left(t\right)=-16\times 2t+64t^0\\ \Rightarrow s`\left(t\right)=-32t+64\end{array}$

Hence, the equation of rate change of the balloon is $s`\left(t\right)=-32t+64$

(b).

To determine

To calculate: Average rate of change of the balloon after the first three seconds.

Answer to Problem 71E

The average speed is 16 feet per second.

Explanation of Solution

Given information: The displacement s, $s\left(t\right)=-16t^2+64t+80$

Formula used: $\text{Then average speed = }\frac{\text{Total time taken}}{\text{Total distance travelled}}$

Calculation:

Total distance after three second:

  $\begin{array}{l}s\left(4\right)=-16{\left(4\right)}^2+64\times 4+80\\ =-16\times 16+256+80\\ =-256+256+80\\ =80\text{ feet}\end{array}$

Total time, when $\text{s}\left(t\right)=0$

Therefore, $s\left(t\right)=-16t^2+64t+80=0$

  $\begin{array}{l}\Rightarrow -16t^2+64t+80=0\\ \Rightarrow -16\left(t^2-4t-5\right)=0\\ \Rightarrow t^2-4t-5=0\\ \Rightarrow t^2-5t+t-5=0\\ \Rightarrow t\left(t-5\right)+1\left(t-5\right)=\\ \Rightarrow \left(t-5\right)\left(t+1\right)=0\\ \therefore \begin{array}{cc}\begin{array}{l}\left(t-5\right)=0\\ t=5\end{array}& \begin{array}{l}\left(t+1\right)=0\\ t=-1\end{array}\end{array}\end{array}$

Hence, total time, $t=5\text{ second}$

Therefore,

  $\begin{array}{l}\text{Then average speed = }\frac{\text{Total time taken}}{\text{Total distance travelled}}\\ =\frac{80\text{ feet}}{5\text{ sec}}\\ =16\raisebox{1ex}{$\text{feet}$}\!\left/ \!\raisebox{-1ex}{$\text{sec}$}\right.\end{array}$

Hence, the average speed after three second is 16 feet per second.

(c).

To determine

To calculate: Time in second at which the balloon reaches its maximum height.

Answer to Problem 71E

Balloon reaches maximum height at  $t=2$  seconds.

Explanation of Solution

Given information: The displacement s, $s\left(t\right)=-16t^2+64t+80$

The given equation is similar as $f\left(x\right)=a{x}^2+bx+c$ , since both are parabola.

Now, $a$ is negative, such that $\left(a<0\right)$ , it is case of maxima (maximum)

  

Hence, $a=-16,b=64,\text{ and }c=80$

Formula used: $t=-\frac{b}{2a}$ (The axis of symmetry has the equation)

Calculation:

Put, the value of a and b , we get

  $\begin{array}{l}\Rightarrow t=-\frac{64}{2\times \left(-16\right)}\\ =\frac{-64}{-32}\\ =2\\ \therefore t=2\text{ second}\end{array}$

Hence, at $t=2$ the balloon reached its maximum position.

(d).

To determine

To calculate: The velocity of the balloon as it hit the ground.

Answer to Problem 71E

The velocity of the balloon as it hit the ground is − 96 feet per second.

Explanation of Solution

Given information: The displacement s, $s\left(t\right)=-16t^2+64t+80$

Ball will impact the ground, when $s\left(t\right)=0$

Therefore, $s\left(t\right)=-16t^2+64t+80=0$

Calculation:

  $\begin{array}{l}\Rightarrow -16t^2+64t+80=0\\ \Rightarrow -16\left(t^2-4t-5\right)=0\\ \Rightarrow t^2-4t-5=0\\ \Rightarrow t^2-5t+t-5=0\\ \Rightarrow t\left(t-5\right)+1\left(t-5\right)=\\ \Rightarrow \left(t-5\right)\left(t+1\right)=0\\ \therefore \begin{array}{cc}\begin{array}{l}\left(t-5\right)=0\\ t=5\end{array}& \begin{array}{l}\left(t+1\right)=0\\ t=-1\end{array}\end{array}\end{array}$

Hence, $t=5\text{ second}$

Formula used: $\frac{d}{dx}\left(x^n\right)=n{x}^{n+1}$

Differentiate the equation of displacement ( s ) with respect to $\text{'}t\text{'}$

  $\begin{array}{l}\frac{d}{dt}s=-16\left(2t^{2-1}\right)+64\left(t^{1-1}\right)+0\\ \Rightarrow s`\left(t\right)=-16\times 2t+64t^0\\ \Rightarrow s`\left(t\right)=-32t+64\end{array}$

Put, $t=5\text{ second}$ in above equation, we get

  $\begin{array}{l}\Rightarrow s`\left(t\right)=-32\times 5+64\\ =-160+64\\ =-96\\ \therefore s`\left(t\right)=-96\raisebox{1ex}{$\text{feet}$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.\end{array}$

Hence, velocity at time of impact is − 96 feet per second.

(e).

To determine

To draw: a graph to verify the result of parts (a) and (b)

Answer to Problem 71E

The required graph:

  

Explanation of Solution

Given information: The displacement s, $s\left(t\right)=-16t^2+64t+80$

The above quadratic equation has two solutions one is negative and the second one is positive and equal to 5 seconds.

Therefore, it takes 5 seconds of the object to hit the ground after it has been thrown upward. The graphical meaning to the answers to parts a and d are shown below.

If $t=0$

then, $s\left(0\right)=0$

Calculation:

Hence,

  $\begin{array}{l}s\left(0\right)=-16\times 0+64\times 0+80\\ =80\\ \therefore s\left(0\right)=80\end{array}$

It means the height of the ball from the ground just before throwing, since the initial is 80 feet.