Chapter 11.1, Problem 55E

(a)

To determine

To calculate: The length of the path traveled by the projectile.

Answer to Problem 55E

The required length of the path traveled by the projectile is 840.421 feet.

Explanation of Solution

Given information:

The parametric equations: $x=\left(v_0\cos \theta \right)t$ , $y=\left(v_0\sin \theta \right)t-16t^2$

The angle: $\theta =60°$

The initial velocity: $v_0=150$

Calculation:

The given parametric equations are $x=\left(v_0\cos \theta \right)t$ and $y=\left(v_0\sin \theta \right)t-16t^2$

Differentiate the parametric equations with respect to t .

  $\frac{dx}{dt}=v_0\cos \theta$ , $\frac{dy}{dt}=v_0\sin \theta -32t$

At t = 0; $y=0$

Substitute 0 for y in the expression $y=\left(v_0\sin \theta \right)t-16t^2$ .

  $\begin{array}{c}0=\left(v_0\sin \theta \right)t-16t^2\\ t\left(v_0\sin \theta -16t\right)=0\end{array}$

Thus, $t=0$

And, $\begin{array}{c}{v}_0\sin \theta -16t=0\\ 16t=v_0\sin \theta \\ t=\frac{v_0\sin \theta }{t}\end{array}$

Consider that L be the length of the parametric curve.

So, $L={\displaystyle \underset{t_2}{\overset{t_2}{\int }}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}}dt$

Write the length of the path traveled by the projectile as:

  $L={\displaystyle \underset{0}{\overset{\frac{v_0\sin \theta }{16}}{\int }}\sqrt{{\left(v_0\cos \theta \right)}^2+\left(v_0\sin \theta -32t^2\right)}dt}$

Also,

  $\begin{array}{c}\frac{dy}{dt}=0\\ v_0\sin \theta -32t=0\\ t=\frac{v_0\sin \theta }{32}\end{array}$

The maximum height at $t=\frac{v_0\sin \theta }{32}$ is:

  $y_{\text{max}}=\left(v_0\sin \theta \right)\left(\frac{v_0\sin \theta }{32}\right)-16{\left(\frac{v_0\sin \theta }{32}\right)}^2$

Now, substitute $60°$ for $\theta$ and 150 for $v_0$ in the expression $L={\displaystyle \underset{0}{\overset{\frac{v_0\sin \theta }{16}}{\int }}\sqrt{{\left(v_0\cos \theta \right)}^2+\left(v_0\sin \theta -32t^2\right)}dt}$ .

  $\begin{array}{c}L={\displaystyle \underset{0}{\overset{\frac{75\sin 60°}{8}}{\int }}\sqrt{{\left(150\cos 60°\right)}^2+\left(150\sin 60°-32t^2\right)}dt}\\ =840.421\end{array}$

Hence, the required length of the path traveled by the projectile is 840.421 feet.

(b)

To determine

To calculate: The maximum height of the projectile.

Answer to Problem 55E

The required maximum height of the projectile is 263.672 feet.

Explanation of Solution

Given information:

The parametric equations: $x=\left(v_0\cos \theta \right)t$ , $y=\left(v_0\sin \theta \right)t-16t^2$

The angle: $\theta =60°$

The initial velocity: $v_0=150$

Calculation:

The given parametric equations are $x=\left(v_0\cos \theta \right)t$ and $y=\left(v_0\sin \theta \right)t-16t^2$

Differentiate the parametric equations with respect to t .

  $\frac{dx}{dt}=v_0\cos \theta$ , $\frac{dy}{dt}=v_0\sin \theta -32t$

At t = 0; $y=0$

Substitute 0 for y in the expression $y=\left(v_0\sin \theta \right)t-16t^2$ .

  $\begin{array}{c}0=\left(v_0\sin \theta \right)t-16t^2\\ t\left(v_0\sin \theta -16t\right)=0\end{array}$

Thus, $t=0$

And, $\begin{array}{c}{v}_0\sin \theta -16t=0\\ 16t=v_0\sin \theta \\ t=\frac{v_0\sin \theta }{t}\end{array}$

Consider that L be the length of the parametric curve.

So, $L={\displaystyle \underset{t_2}{\overset{t_2}{\int }}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}}dt$

Write the length of the path traveled by the projectile as:

  $L={\displaystyle \underset{0}{\overset{\frac{v_0\sin \theta }{16}}{\int }}\sqrt{{\left(v_0\cos \theta \right)}^2+\left(v_0\sin \theta -32t^2\right)}dt}$

Also,

  $\begin{array}{c}\frac{dy}{dt}=0\\ v_0\sin \theta -32t=0\\ t=\frac{v_0\sin \theta }{32}\end{array}$

The maximum height at $t=\frac{v_0\sin \theta }{32}$ is:

  $y_{\text{max}}=\left(v_0\sin \theta \right)\left(\frac{v_0\sin \theta }{32}\right)-16{\left(\frac{v_0\sin \theta }{32}\right)}^2$

Substitute $60°$ for $\theta$ and $150$ for $v_0$ in the expression $y_{\text{max}}=\left(v_0\sin \theta \right)\left(\frac{v_0\sin \theta }{32}\right)-16{\left(\frac{v_0\sin \theta }{32}\right)}^2$ .

  $\begin{array}{c}{y}_{\text{max}}=\left(150\sin 60°\right)\left(\frac{150\sin 60°}{32}\right)-16{\left(\frac{150\sin 60°}{32}\right)}^2\\ =263.672\end{array}$

Hence, the required maximum height of the projectile is 263.672 feet.