Chapter 10.1, Problem 54E

$\text{(a)}$

To determine

To find: The length of the path.

Answer to Problem 54E

The answer: $L\approx 641.236\text{ft}$

Explanation of Solution

Given information:

The parametric equations are:

  $\begin{array}{c}x=\left(v_0\cos \theta \right)t,\\ y=\left(v_0\sin \theta \right)t-16t^2\end{array}$

And,

  $\begin{array}{c}\theta ={30}^{\circ },\\ v_0=150\end{array}$

Calculation:

In this exercise,

  $\begin{array}{c}\theta ={30}^{\circ },\\ v_0=150\end{array}$

The parametric equations for the projectile's trajectory are then equal to:

  $\begin{array}{c}x=\left(150\cos {30}^{\circ }\right)t\\ =150\cdot \frac{\sqrt{3}}{2}\cdot t\\ =75\sqrt{3}t\\ y=\left(150\sin {30}^{\circ }\right)t-16t^2\\ =150\cdot \frac{1}{2}\cdot t-16t^2\\ =75t-16t^2\end{array}$

To use the previous result, then,

  $\begin{array}{c}75-16t=0\\ \iff 16t=75\\ \iff t=\frac{75}{16}\end{array}$

Know that,

  

In this exercise,

  $\begin{array}{c}x=75\sqrt{3}t\\ y=75t-16t^2\end{array}$

Then,

  $\begin{array}{c}\frac{dx}{dt}=75\sqrt{3}\\ \frac{dy}{dt}=75-32t\end{array}$

To use the prior findings, determine that the projectile's route length is equal to:

  

Therefore, the required length of the path is $L\approx 641.236\text{ft}$ .

$\text{(b)}$

To determine

To find: The maximum height of the projectile.

Answer to Problem 54E

The answer: $\approx 87.891\text{ft}$

Explanation of Solution

Given information:

The parametric equations are:

  $\begin{array}{c}x=\left(v_0\cos \theta \right)t,\\ y=\left(v_0\sin \theta \right)t-16t^2\end{array}$

And,

  $\begin{array}{c}\theta ={30}^{\circ },\\ v_0=150\end{array}$

Calculation:

In this exercise,

  $\begin{array}{c}\theta ={30}^{\circ },\\ v_0=150\end{array}$

The parametric equations for the projectile's trajectory are then equal to:

  $\begin{array}{c}x=\left(150\cos {30}^{\circ }\right)t\\ =150\cdot \frac{\sqrt{3}}{2}\cdot t\\ =75\sqrt{3}t\\ y=\left(150\sin {30}^{\circ }\right)t-16t^2\\ =150\cdot \frac{1}{2}\cdot t-16t^2\\ =75t-16t^2\end{array}$

To estimate the maximum height of the path of the projectile, to find the maximum of the function $y=75t-16t^2$

Know that,

  $\frac{dy}{dt}\left(t_0\right)=0$

To use the previous result,

  $\begin{array}{c}75-32t=0\\ \iff 32t=75\\ \iff t=\frac{75}{32}\end{array}$

The previous findings indicate that the projectile's greatest height along its route is equal to:

  $\begin{array}{c}y(75/32)=75\cdot \frac{75}{32}-16\cdot {\left(\frac{75}{32}\right)}^2\\ =\frac{{75}^2}{32}-\frac{1}{2}\cdot \frac{{75}^2}{32}\\ =\frac{{75}^2}{64}\\ =\frac{5625}{64}\\ \approx 87.891\text{ft}\end{array}$

Therefore, the required maximum height of the projectile is $\approx 87.891\text{ft}$ .