Chapter 2, Problem 2.22P

(a)

To determine

The maximum range in vacuum.

Answer to Problem 2.22P

The maximum range in vacuum is $\underset{\_}{1}$.

Explanation of Solution

Write the expression for the range of the projectile.

  $R_{\max }=\frac{v_0}{g}$        (I)

Here, $v_0$ is the initial velocity, and $g$ is the acceleration due to gravity.

Conclusion:

Substitute, $1$ for $v_0$, and $g$ in equation (I) to find the maximum range.

  $\begin{array}{c}{R}_{\max }=\frac{1}{1}\\ =1\end{array}$

Therefore, the maximum range in vacuum is $\underset{\_}{1}$.

(b)

To determine

The range in a given medium at an angle $\theta =0.75$.

Answer to Problem 2.22P

The range in a given medium at an angle $\theta =0.75$ is $\underset{\_}{0.491094}$.

Explanation of Solution

Consider the linear drag.

The range can be solved from the following equation.

  $\frac{v_{y0}+v_{ter}}{v_{x0}}R+v_{ter}\tau \ln \left(1-\frac{R}{v_{x0}\tau }\right)=0$        (II)

Here, $v_{ter}$ is the terminal velocity, $v_{x0}$, and $v_{y0}$ be the horizontal and vertical component of initial velocity.

Conclusion:

Substitute, $1$ for $v_0$, $1$ for $v_{ter}$, and $g$, $\frac{v_{ter}}{g}$ for $\tau$, $v_0\cos \theta$ for $v_{x0}$, $v_0\sin \theta$ for $v_{y0}$, and $\frac{\pi }{4}$ for $\theta$ in equation (I).

  $\begin{array}{c}\frac{v_0\sin \theta +v_{ter}}{v_0\cos \theta }R+v_{ter}\left(\frac{v_{ter}}{g}\right)\ln \left(1-\frac{R}{v_0\cos \theta \left(\frac{v_{ter}}{g}\right)}\right)=0\\ \frac{\left(1\right)\sin \left(\frac{\pi }{4}\right)+1}{\left(1\right)\cos \left(\frac{\pi }{4}\right)}R+\left(1\right)\left(\frac{1}{1}\right)\ln \left(1-\frac{R}{v_0\cos \left(\frac{\pi }{4}\right)\left(\frac{1}{1}\right)}\right)=0\\ \frac{0.7071+1}{0.7071}R+\ln \left(1-\frac{R}{\left(0.7071\right)\left(1\right)}\right)=0\\ 2.414R+\ln \left(101.414R\right)=0\end{array}$        (III)

Solve equation (III) to obtain the value of range, $R$.

  $\begin{array}{c}2.414R+\ln \left(101.414R\right)=0\\ R=0.491094\end{array}$

Therefore, the range in a given medium at an angle $\theta =0.75$ is $\underset{\_}{0.491094}$.

(c)

To determine

The range in a given medium for some selected values of $\theta =0.4,0.5,0.6,0.7,0.8$.

Answer to Problem 2.22P

The range in a given medium at an angle $\theta =0.4$ is $\underset{\_}{0.462875}$, $\theta =0.5$ is $\underset{\_}{0.499439}$, $\theta =0.6$ is $\underset{\_}{0.513562}$, $\theta =0.7$ is $\underset{\_}{0.508474}$, $\theta =0.8$ is $\underset{\_}{0.486909}$.

Explanation of Solution

Use equation (II) to solve range for different values of $\theta$.

Conclusion:

Substitute, $1$ for $v_0$, $1$ for $v_{ter}$, and $g$, $\frac{v_{ter}}{g}$ for $\tau$, $v_0\cos \theta$ for $v_{x0}$, $v_0\sin \theta$ for $v_{y0}$, and $0.4$ for $\theta$ in equation (II).

  $\begin{array}{c}\frac{v_0\sin \theta +v_{ter}}{v_0\cos \theta }R+v_{ter}\left(\frac{v_{ter}}{g}\right)\ln \left(1-\frac{R}{v_0\cos \left(0.4\right)\left(\frac{v_{ter}}{g}\right)}\right)=0\\ \frac{\left(1\right)\sin \left(0.4\right)+1}{\left(1\right)\cos \left(0.4\right)}R+\left(1\right)\left(\frac{1}{1}\right)\ln \left(1-\frac{R}{v_0\cos \left(0.4\right)\left(\frac{1}{1}\right)}\right)=0\\ \frac{0.3894+1}{0.92106}R+\ln \left(1-\frac{R}{\left(0.92106\right)\left(1\right)}\right)=0\\ 1.508R+\ln \left(1-1.0857R\right)=0\end{array}$        (IV)

Solve equation (IV) to obtain the value of range, $R$.

  $\begin{array}{c}1.508R+\ln \left(1-1.0857R\right)=0\\ R=0.462875\end{array}$

Substitute, $1$ for $v_0$, and $g$, $\frac{v_{ter}}{g}$ for $\tau$, $v_0\cos \theta$ for $v_{x0}$, $v_0\sin \theta$ for $v_{y0}$, and $0.5$ for $\theta$ in equation (II).

  $\begin{array}{c}\frac{v_0\sin \theta +v_{ter}}{v_0\cos \theta }R+v_{ter}\left(\frac{v_{ter}}{g}\right)\ln \left(1-\frac{R}{v_0\cos \left(0.4\right)\left(\frac{v_{ter}}{g}\right)}\right)=0\\ \frac{\left(1\right)\sin \left(0.5\right)+1}{\left(1\right)\cos \left(0.5\right)}R+\left(1\right)\left(\frac{1}{1}\right)\ln \left(1-\frac{R}{v_0\cos \left(0.5\right)\left(\frac{1}{1}\right)}\right)=0\\ \frac{0.4794+1}{0.8776}R+\ln \left(1-\frac{R}{\left(0.8776\right)\left(1\right)}\right)=0\\ 1.6857R+\ln \left(1-1.1395R\right)=0\end{array}$        (V)

Solve equation (V) to obtain the value of range, $R$.

  $\begin{array}{c}1.6857R+\ln \left(1-1.1395R\right)=0\\ R=0.499439\end{array}$

Substitute, $1$ for $v_0$, and $g$, $\frac{v_{ter}}{g}$ for $\tau$, $v_0\cos \theta$ for $v_{x0}$, $v_0\sin \theta$ for $v_{y0}$, and $0.6$ for $\theta$ in equation (II).

  $\begin{array}{c}\frac{v_0\sin \theta +v_{ter}}{v_0\cos \theta }R+v_{ter}\left(\frac{v_{ter}}{g}\right)\ln \left(1-\frac{R}{v_0\cos \left(0.4\right)\left(\frac{v_{ter}}{g}\right)}\right)=0\\ \frac{\left(1\right)\sin \left(0.6\right)+1}{\left(1\right)\cos \left(0.6\right)}R+\left(1\right)\left(\frac{1}{1}\right)\ln \left(1-\frac{R}{\left(1\right)\cos \left(0.6\right)\left(\frac{1}{1}\right)}\right)=0\\ \frac{0.5645+1}{0.8253}R+\ln \left(1-\frac{R}{\left(0.8253\right)\left(1\right)}\right)=0\\ 1.8958R+\ln \left(1-1.2117R\right)=0\end{array}$        (VI)

Solve equation (VI) to obtain the value of range, $R$.

  $\begin{array}{c}1.8958R+\ln \left(1-1.2117R\right)=0\\ R=0.513562\end{array}$

Substitute, $1$ for $v_0$, and $g$, $\frac{v_{ter}}{g}$ for $\tau$, $v_0\cos \theta$ for $v_{x0}$, $v_0\sin \theta$ for $v_{y0}$, and $0.7$ for $\theta$ in equation (II).

  $\begin{array}{c}\frac{v_0\sin \theta +v_{ter}}{v_0\cos \theta }R+v_{ter}\left(\frac{v_{ter}}{g}\right)\ln \left(1-\frac{R}{v_0\cos \left(0.4\right)\left(\frac{v_{ter}}{g}\right)}\right)=0\\ \frac{\left(1\right)\sin \left(0.7\right)+1}{\left(1\right)\cos \left(0.7\right)}R+\left(1\right)\left(\frac{1}{1}\right)\ln \left(1-\frac{R}{\left(1\right)\cos \left(0.7\right)\left(\frac{1}{1}\right)}\right)=0\\ \frac{0.6442+1}{0.7648}R+\ln \left(1-\frac{R}{\left(0.7648\right)\left(1\right)}\right)=0\\ 2.1498R+\ln \left(1-1.3075R\right)=0\end{array}$        (VII)

Solve equation (VII) to obtain the value of range, $R$.

  $\begin{array}{c}2.1498R+\ln \left(1-1.3075R\right)=0\\ R=0.508474\end{array}$

Substitute, $1$ for $v_0$, and $g$, $\frac{v_{ter}}{g}$ for $\tau$, $v_0\cos \theta$ for $v_{x0}$, $v_0\sin \theta$ for $v_{y0}$, and $0.8$ for $\theta$ in equation (II).

  $\begin{array}{c}\frac{v_0\sin \theta +v_{ter}}{v_0\cos \theta }R+v_{ter}\left(\frac{v_{ter}}{g}\right)\ln \left(1-\frac{R}{v_0\cos \left(0.4\right)\left(\frac{v_{ter}}{g}\right)}\right)=0\\ \frac{\left(1\right)\sin \left(0.8\right)+1}{\left(1\right)\cos \left(0.8\right)}R+\left(1\right)\left(\frac{1}{1}\right)\ln \left(1-\frac{R}{\left(1\right)\cos \left(0.8\right)\left(\frac{1}{1}\right)}\right)=0\\ \frac{0.7173+1}{0.6967}R+\ln \left(1-\frac{R}{\left(0.6967\right)\left(1\right)}\right)=0\\ 2.4649R+\ln \left(1-1.4353R\right)=0\end{array}$        (VIII)

Solve equation (VIII) to obtain the value of range, $R$.

  $\begin{array}{c}2.4649R+\ln \left(1-1.4353R\right)=0\\ R=0.486909\end{array}$

Therefore, the range in a given medium at an angle $\theta =0.4$ is $\underset{\_}{0.462875}$, $\theta =0.5$ is $\underset{\_}{0.499439}$, $\theta =0.6$ is $\underset{\_}{0.513562}$, $\theta =0.7$ is $\underset{\_}{0.508474}$, $\theta =0.8$ is $\underset{\_}{0.486909}$.

(d)

To determine

To calculate range corresponding to smaller interval until the maximum range is obtained.

Answer to Problem 2.22P

The maximum range is $\underset{\_}{0.514043}$ at an angle $\underset{\_}{\theta =0.62}$.

Explanation of Solution

Use equation (II) to find the range.

Conclusion:

Substitute, $1$ for $v_0$, and $g$, $\frac{v_{ter}}{g}$ for $\tau$, $v_0\cos \theta$ for $v_{x0}$, $v_0\sin \theta$ for $v_{y0}$, and $0.62$ for $\theta$ in equation (II.

  $\begin{array}{c}\frac{v_0\sin \theta +v_{ter}}{v_0\cos \theta }R+v_{ter}\left(\frac{v_{ter}}{g}\right)\ln \left(1-\frac{R}{v_0\cos \theta \left(\frac{v_{ter}}{g}\right)}\right)=0\\ \frac{\left(1\right)\sin \left(0.62\right)+1}{\left(1\right)\cos \left(0.62\right)}R+\left(1\right)\left(\frac{1}{1}\right)\ln \left(1-\frac{R}{v_0\cos \left(0.62\right)\left(\frac{1}{1}\right)}\right)=0\\ \frac{0.5810+1}{0.8139}R+\ln \left(1-\frac{R}{\left(0.8139\right)\left(1\right)}\right)=0\\ 1.9425R+\ln \left(1-1.22865R\right)=0\end{array}$        (IX)

Solve equation (IX) to obtain the value of range, $R$.

  $\begin{array}{c}1.9425R+\ln \left(1-1.22865R\right)=0\\ R=0.514043\end{array}$

Substitute, $1$ for $v_0$, and $g$, $\frac{v_{ter}}{g}$ for $\tau$, $v_0\cos \theta$ for $v_{x0}$, $v_0\sin \theta$ for $v_{y0}$, and $0.62$ for $\theta$ in equation (II).

  $\begin{array}{c}\frac{v_0\sin \theta +v_{ter}}{v_0\cos \theta }R+v_{ter}\left(\frac{v_{ter}}{g}\right)\ln \left(1-\frac{R}{v_0\cos \theta \left(\frac{v_{ter}}{g}\right)}\right)=0\\ \frac{\left(1\right)\sin \left(0.62\right)+1}{\left(1\right)\cos \left(0.62\right)}R+\left(1\right)\left(\frac{1}{1}\right)\ln \left(1-\frac{R}{v_0\cos \left(0.62\right)\left(\frac{1}{1}\right)}\right)=0\\ \frac{0.5810+1}{0.8139}R+\ln \left(1-\frac{R}{\left(0.8139\right)\left(1\right)}\right)=0\\ 1.9425R+\ln \left(1-1.22865R\right)=0\end{array}$        (IX)

Solve equation (IX) to obtain the value of range, $R$.

$\begin{array}{c}1.9425R+\ln \left(1-1.22865R\right)=0\\ R=0.514043\end{array}$

Substitute, $1$ for $v_0$, and $g$, $\frac{v_{ter}}{g}$ for $\tau$, $v_0\cos \theta$ for $v_{x0}$, $v_0\sin \theta$ for $v_{y0}$, and $0.63$ for $\theta$ in equation (II).

  $\begin{array}{c}\frac{v_0\sin \theta +v_{ter}}{v_0\cos \theta }R+v_{ter}\left(\frac{v_{ter}}{g}\right)\ln \left(1-\frac{R}{v_0\cos \theta \left(\frac{v_{ter}}{g}\right)}\right)=0\\ \frac{\left(1\right)\sin \left(0.63\right)+1}{\left(1\right)\cos \left(0.63\right)}R+\left(1\right)\left(\frac{1}{1}\right)\ln \left(1-\frac{R}{v_0\cos \left(0.63\right)\left(\frac{1}{1}\right)}\right)=0\\ \frac{0.5891+1}{0.8080}R+\ln \left(1-\frac{R}{\left(0.8080\right)\left(1\right)}\right)=0\\ 1.9667R+\ln \left(1-1.2376R\right)=0\end{array}$        (X)

Solve equation (X) to obtain the value of range, $R$.

$\begin{array}{c}1.9667R+\ln \left(1-1.2376R\right)=0\\ R=0.513955\end{array}$

Substitute, $1$ for $v_0$, and $g$, $\frac{v_{ter}}{g}$ for $\tau$, $v_0\cos \theta$ for $v_{x0}$, $v_0\sin \theta$ for $v_{y0}$, and $0.64$ for $\theta$ in equation (II).

  $\begin{array}{c}\frac{v_0\sin \theta +v_{ter}}{v_0\cos \theta }R+v_{ter}\left(\frac{v_{ter}}{g}\right)\ln \left(1-\frac{R}{v_0\cos \theta \left(\frac{v_{ter}}{g}\right)}\right)=0\\ \frac{\left(1\right)\sin \left(0.64\right)+1}{\left(1\right)\cos \left(0.64\right)}R+\left(1\right)\left(\frac{1}{1}\right)\ln \left(1-\frac{R}{v_0\cos \left(0.64\right)\left(\frac{1}{1}\right)}\right)=0\\ \frac{0.5972+1}{0.8021}R+\ln \left(1-\frac{R}{\left(0.8021\right)\left(1\right)}\right)=0\\ 1.9913R+\ln \left(1-1.2467R\right)=0\end{array}$        (XI)

Solve equation (XI) to obtain the value of range, $R$.

$\begin{array}{c}1.9913R+\ln \left(1-1.2467R\right)=0\\ R=0.513758\end{array}$

Thus from the above calculations the value of $R$ increases up to $\theta =0.62$ and then decreases.

Therefore, the maximum range is $\underset{\_}{0.514043}$  at an angle $\underset{\_}{\theta =0.62}$.