Chapter 10.1, Problem 55E

$\text{(a)}$

To determine

To find: The length of the path.

Answer to Problem 55E

The answer: $L\approx 840.421\text{ft}$

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,

  $\begin{array}{c}\theta ={60}^{\circ },\\ v_0=150\end{array}$

Calculation:

In this exercise,

  $\begin{array}{c}\theta ={60}^{\circ },\\ v_0=150\end{array}$

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

  $\begin{array}{c}x=\left(150\cos {60}^{\circ }\right)t\\ =150\cdot \frac{1}{2}\cdot t=75t,\\ y=\left(150\sin {60}^{\circ }\right)t-16t^2\\ =150\cdot \frac{\sqrt{3}}{2}\cdot t-16t^2\\ =75\sqrt{3}t-16t^2\end{array}$

To use the previous result, then,

  $\begin{array}{c}75\sqrt{3}-16t=0\\ \iff 16t=75\sqrt{3}\\ \iff t=\frac{75\sqrt{3}}{16}\end{array}$

Know that,

  

In this exercise,

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

Then,

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

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

  

Therefore, the required length of the path is $L\approx 840.421\text{ft}$ .

$\text{(b)}$

To determine

To find: The maximum height of the projectile.

Answer to Problem 55E

The answer: $\approx 263.672\text{ft}$

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,

  $\begin{array}{c}\theta ={60}^{\circ },\\ v_0=150\end{array}$

Calculation:

In this exercise,

  $\begin{array}{c}\theta ={60}^{\circ },\\ v_0=150\end{array}$

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

  $\begin{array}{c}x=\left(150\cos {60}^{\circ }\right)t\\ =150\cdot \frac{1}{2}\cdot t=75t,\\ y=\left(150\sin {60}^{\circ }\right)t-16t^2\\ =150\cdot \frac{\sqrt{3}}{2}\cdot t-16t^2\\ =75\sqrt{3}t-16t^2\end{array}$

To estimate the maximum height of the path of the projectile, to find the maximum of the function $y=75\sqrt{3}t-16t^2$ .

Know that,

  $\frac{dy}{dt}\left(t_0\right)=0$

To use the previous result,

  $\begin{array}{c}75\sqrt{3}-32t=0\\ \iff 32t=75\sqrt{3}\\ \iff t=\frac{75\sqrt{3}}{32}\end{array}$

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

  $\begin{array}{c}y(75\sqrt{3}/32)=75\sqrt{3}\cdot \frac{75\sqrt{3}}{32}-16\cdot {\left(\frac{75\sqrt{3}}{32}\right)}^2\\ =\frac{{(75\sqrt{3})}^2}{32}-\frac{1}{2}\cdot \frac{{(75\sqrt{3})}^2}{32}\\ =\frac{{(75\sqrt{3})}^2}{64}\\ =\frac{16,875}{64}\\ \approx 263.672\text{ft}\end{array}$

Therefore, the required maximum height of the projectile is $\approx 263.672\text{ft}$ .