Chapter 10.1, Problem 56E

$\text{(a)}$

To determine

To find: The length of the path.

Answer to Problem 56E

The answer: $L=703.125\text{ft}$

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,

  $\begin{array}{c}\theta ={90}^{\circ },\\ v_0=150\end{array}$

Calculation:

In this exercise,

  $\begin{array}{c}\theta ={90}^{\circ },\\ v_0=150\end{array}$

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

  $\begin{array}{c}x=\left(150\cos {90}^{\circ }\right)t\\ =150\cdot 0\cdot t\\ =0\\ y=\left(150\sin {90}^{\circ }\right)t-16t^2\\ =150\cdot 1\cdot t-16t^2\\ =150t-16t^2\end{array}$

To use the previous result, then,

  $\begin{array}{c}150-16t=0\\ \iff 16t=150\\ \iff t=\frac{150}{16}\\ =\frac{75}{8}\end{array}$

Know that,

  

In this exercise,

  $\begin{array}{c}x=0\\ y=150t-16t^2\end{array}$

Then,

  $\begin{array}{c}\frac{dx}{dt}=0\\ \frac{dy}{dt}=150-32t\end{array}$

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

  

Therefore, the required length of the path is $L=703.125\text{ft}$ .

$\text{(b)}$

To determine

To find: The maximum height of the projectile.

Answer to Problem 56E

The answer: $351.5625\text{ft}$

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,

  $\begin{array}{c}\theta ={90}^{\circ },\\ v_0=150\end{array}$

Calculation:

In this exercise,

  $\begin{array}{c}\theta ={90}^{\circ },\\ v_0=150\end{array}$

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

  $\begin{array}{c}x=\left(150\cos {90}^{\circ }\right)t\\ =150\cdot 0\cdot t\\ =0\\ y=\left(150\sin {90}^{\circ }\right)t-16t^2\\ =150\cdot 1\cdot t-16t^2\\ =150t-16t^2\end{array}$

To estimate the maximum height of the path of the projectile, to find the maximum of the function $y=150t-16t^2$ .

Know that,

  $\frac{dy}{dt}\left(t_0\right)=0$

To use the previous result,

  $\begin{array}{c}150-32t=0\\ \iff 32t=150\\ \iff t=\frac{150}{32}\\ =\frac{75}{16}\end{array}$

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

  $\begin{array}{c}y(75/16)=150\cdot \frac{75}{16}-16\cdot {\left(\frac{75}{16}\right)}^2\\ =\frac{{75}^2}{8}-\frac{{75}^2}{16}\\ =\frac{{75}^2}{16}\\ =\frac{5625}{16}\\ =351.5625\text{ft}\end{array}$

Therefore, the required maximum height of the projectile is $351.5625\text{ft}$ .