Chapter 10.1, Problem 53E

$\text{(a)}$

To determine

To find: The length of the path.

Answer to Problem 53E

The answer: $L\approx 461.746\text{ft}$

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,

  $\begin{array}{c}\theta ={20}^{\circ },\\ v_0=150\end{array}$

Calculation:

In this exercise,

  $\begin{array}{c}\theta ={20}^{\circ },\\ v_0=150\end{array}$

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

  $\begin{array}{c}x=\left(150\cos {20}^{\circ }\right)t\\ y=\left(150\sin {20}^{\circ }\right)t-16t^2\end{array}$

To use the previous result, then,

  $\begin{array}{c}150\sin {20}^{\circ }-16t=0\\ \iff 16t=150\sin {20}^{\circ }\\ \iff t=\frac{150\sin {20}^{\circ }}{16}\\ =\frac{75\sin {20}^{\circ }}{8}\end{array}$

Know that,

  

In this exercise,

  $\begin{array}{c}x=\left(150\cos {20}^{\circ }\right)t\\ y=\left(150\sin {20}^{\circ }\right)t-16t^2\end{array}$

Then,

  $\begin{array}{c}\frac{dx}{dt}=150\cos {20}^{\circ }\\ \frac{dy}{dt}=150\sin {20}^{\circ }-32t\end{array}$

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

  

Therefore, the required length of the path is $L\approx 461.746\text{ft}$ .

$\text{(b)}$

To determine

To find: The maximum height of the projectile.

Answer to Problem 53E

The answer: $\approx 41.125\text{ft}$

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,

  $\begin{array}{c}\theta ={20}^{\circ },\\ v_0=150\end{array}$

Calculation:

In this exercise,

  $\begin{array}{c}\theta ={20}^{\circ },\\ v_0=150\end{array}$

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

  $\begin{array}{c}x=\left(150\cos {20}^{\circ }\right)t\\ y=\left(150\sin {20}^{\circ }\right)t-16t^2\end{array}$

To estimate the maximum height of the path of the projectile, to find the maximum of the function $y=\left(150\sin {20}^{\circ }\right)t-16t^2$ .

Know that,

  $\frac{dy}{dt}\left(t_0\right)=0$

To use the previous result,

  $\begin{array}{c}150\sin {20}^{\circ }-32t=0\\ \iff 32t=150\sin {20}^{\circ }\\ \iff t=\frac{150\sin {20}^{\circ }}{32}\\ =\frac{75\sin {20}^{\circ }}{16}\end{array}$

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

  $\begin{array}{c}y\left(\left(75\sin {20}^{\circ }\right)/16\right)=150\sin {20}^{\circ }\cdot \frac{75\sin {20}^{\circ }}{16}-16\cdot {\left(\frac{75\sin {20}^{\circ }}{16}\right)}^2\\ =\frac{{\left(75\sin {20}^{\circ }\right)}^2}{8}-\frac{{\left(75\sin {20}^{\circ }\right)}^2}{16}\\ =\frac{{\left(75\sin {20}^{\circ }\right)}^2}{16}\\ \approx 41.125\text{ft}\end{array}$

Therefore, the required maximum height of the projectile is $\approx 41.125\text{ft}$ .