Chapter 2, Problem 2.35P

(a)

To determine

The equations for the velocity and position of a dropped object subjected to quadratic air resistance.

Answer to Problem 2.35P

The equations for the velocity and position of a dropped object subjected to quadratic air resistance are $\underset{\_}{v=v_{ter}\tanh \left(\frac{gt}{v_{ter}}\right)}$, and $\underset{\_}{y=\frac{{\left(v_{ter}\right)}^2}{g}\ln \left[\cos h\left(\frac{gt}{v_{ter}}\right)\right]}$.

Explanation of Solution

Write the equation of motion of a baseball dropped from a window in a high tower.

  $m\dot{v}=mg-c{v}^2$        (I)

Here, $m$ is the mass, $g$ is the acceleration due to gravity, $v$ is the velocity of the ball, and $c$ is a constant.

When ball reaches terminal velocity the velocity of the ball will be zero.

Substitute, $0$ for $\dot{v}$ in equation (I).

  $\begin{array}{c}m\left(0\right)=mg-c{v}_{ter}^2\\ c{v}_{ter}^2=mg\\ c=\frac{mg}{v_{ter}^2}\end{array}$

Substitute, $\frac{mg}{v_{ter}^2}$ for $c$ in equation (I).

  $\begin{array}{c}m\dot{v}=mg-\left(\frac{mg}{v_{ter}^2}\right)v^2\\ \dot{v}=g-g\frac{v^2}{v_{ter}^2}\end{array}$        (II)

Rewrite equation (II).

  $\begin{array}{c}\frac{dv}{dt}=g\left(1-\frac{v^2}{v_{ter}^2}\right)\\ \frac{dv}{1-\frac{v^2}{v_{ter}^2}}=gdt\end{array}$        (III)

Integrate equation (III) on both sides to obtain $v$ in terms of $t$.

  ${\displaystyle \underset{0}{\overset{v}{\int }}\frac{d{v}^{\prime }}{1-\frac{v^2}{v_{ter}^2}}}={\displaystyle \underset{0}{\overset{t}{\int }}gd{t}^{\prime }}$        (IV)

Substitute, $t$ for $\frac{v^{\prime }}{v_{ter}}$, and $v_{ter}dt$ for $d{v}^{\prime }$ in equation (IV).

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{v}{\int }}\frac{v_{ter}dt}{1-t^2}}={\displaystyle \underset{0}{\overset{t}{\int }}gd{t}^{\prime }}\\ v_{ter}\tan h^{-1}\left(t\right)=g\left(t-0\right)\\ {\left[v_{ter}\tan h^{-1}\left(\frac{v^{\prime }}{v_{ter}}\right)\right]}_0^v=gt\\ v_{ter}\left[\arctan h^{-1}\left(\frac{v}{v_{ter}}\right)-{\tan }^{-1}h\left(0\right)\right]=gt\end{array}$        (V)

Substitute, $0$ for $\tan h^{-1}\left(0\right)$ in equation (V).

$\begin{array}{c}{v}_{ter}\left[\arctan h^{-1}\left(\frac{v}{v_{ter}}\right)\right]=gt\\ t=\frac{v_{ter}}{g}\arctan \left(\frac{v}{v_{ter}}\right)\\ \frac{v}{v_{ter}}=\tan h\left(\frac{gt}{v_{ter}}\right)\\ v=v_{ter}\tan h\left(\frac{gt}{v_{ter}}\right)\end{array}$        (VI)

Substitute, $\frac{dy}{dt}$ for $v$ in equation (VI).

  $\begin{array}{c}\frac{dy}{dt}=v_{ter}\tan h\left(\frac{gt}{v_{ter}}\right)\\ dy=v_{ter}\tan h\left(\frac{gt}{v_{ter}}\right)dt\end{array}$        (VII)

Integrate equation (VII) on both sides to get $y$.

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{y}{\int }}dy}={\displaystyle \underset{0}{\overset{t}{\int }}{v}_{ter}\tan h\left(\frac{gt}{v_{ter}}\right)dt}\\ {\left(y\right)}_0^y=v_{ter}{\left[\frac{\ln \left[\cosh \left(\frac{g{t}^{\prime }}{v_{ter}}\right)\right]}{\left(\frac{g}{v_{ter}}\right)}\right]}_0^t\\ y-0=\frac{{\left(v_{ter}\right)}^2}{g}\left\{\ln \left[\cosh \left(\frac{gt}{v_{ter}}\right)-\ln \cos h\left(0\right)\right]\right\}\end{array}$        (VIII)

Substitute, $0$ for $\ln \cosh \left(0\right)$ in equation (VIII).

  $y=\frac{{\left(v_{ter}\right)}^2}{g}\ln \left[\cosh \left(\frac{gt}{v_{ter}}\right)\right]$        (IX)

Conclusion:

Therefore, the equations for the velocity and position of a dropped object subjected to quadratic air resistance are $\underset{\_}{v=v_{ter}\tanh \left(\frac{gt}{v_{ter}}\right)}$, and $\underset{\_}{y=\frac{{\left(v_{ter}\right)}^2}{g}\ln \left[\cos h\left(\frac{gt}{v_{ter}}\right)\right]}$.

(b)

To determine

Show that when $t=\tau$ velocity reaches $76\%$ of the terminal speed also the percentage corresponding to $t=2\tau$, and $t=3\tau$.

Answer to Problem 2.35P

It is show that when $t=\tau$ velocity reaches $76\%$ of the terminal speed also the percentage corresponding to $t=2\tau$, is $\underset{\_}{96\%}$, and percentage corresponding to $t=3\tau$ is $\underset{\_}{99.5\%}$.

Explanation of Solution

Substitute, $\frac{v_{ter}}{g}$ for $\tau$ in equation (VI).

  $v=v_{ter}\tan h\left(\frac{t}{\tau }\right)$        (X)

Substitute, $t=\tau$ in equation (X).

  $\begin{array}{c}v=v_{ter}\tan h\left(1\right)\\ =v_{ter}\left(0.761594\right)\\ =0.76v_{ter}\end{array}$

Thus, when $t=\tau$, $v$ has reached $76\%$ of the terminal velocity.

Substitute, $t=2\tau$ in equation (X).

  $\begin{array}{c}v=v_{ter}\tan h\left(2\right)\\ =v_{ter}\left(0.96402\right)\\ =0.96v_{ter}\end{array}$

Thus, when $t=2\tau$, $v$ has reached $96\%$ of the terminal velocity.

Substitute, $t=3\tau$ in equation (X).

  $\begin{array}{c}v=v_{ter}\tan h\left(3\right)\\ =v_{ter}\left(0.99505\right)\\ =0.995v_{ter}\end{array}$

Thus, when $t=3\tau$, $v$ has reached $99.5\%$ of the terminal velocity.

Conclusion:

Therefore, it is show that when $t=\tau$ velocity reaches $76\%$ of the terminal speed also the percentage corresponding to $t=2\tau$, is $\underset{\_}{96\%}$, and percentage corresponding to $t=3\tau$ is $\underset{\_}{99.5\%}$.

(c)

To determine

Show that when $t>>\tau$, the position is $y=v_{ter}t+\text{cons}$.

Answer to Problem 2.35P

It is shown that when $t>>\tau$, the position is $y=v_{ter}t+\text{cons}$.

Explanation of Solution

When $t>>\tau$, then $\frac{t}{\tau }>>1$, $\cosh \left(\frac{t}{\tau }\right)$ will be.

  $\cosh \left(\frac{t}{\tau }\right)=\frac{e^{\frac{t}{\tau }}+e^{-\frac{t}{\tau }}}{2}$

As $x\to \infty$, $e^{-x}\to 0$.

  $\cosh \left(\frac{t}{\tau }\right)\approx \frac{e^{\frac{t}{\tau }}}{2}$

Substitute, $\frac{v_{ter}}{g}$ for $\tau$ in equation (IX), and use the above simplification.

  $\begin{array}{c}y=\frac{{\left(v_{ter}\right)}^2}{g}\ln \left[\cosh \left(\frac{t}{\tau }\right)\right]\\ =\frac{{\left(v_{ter}\right)}^2}{g}\ln \left[\frac{e^{\frac{t}{\tau }}}{2}\right]\\ =\frac{{\left(v_{ter}\right)}^2}{g}\ln \left[\frac{e^{\frac{t}{\tau }}}{2}-\ln 2\right]\\ =\frac{{\left(v_{ter}\right)}^2}{g}\ln \left[\frac{e^{\frac{t}{\tau }}}{2}-\ln 2\right]\end{array}$        (XI)

Simplify equation (XI).

  $\begin{array}{c}y=\frac{{\left(v_{ter}\right)}^2}{g}\ln \left[\frac{t}{\tau }-\ln 2\right]\\ =\frac{{\left(v_{ter}\right)}^2}{g}\ln \left[t\frac{g}{v_{ter}}-\ln 2\right]\\ \approx \frac{{\left(v_{ter}\right)}^2}{g}t\frac{g}{v_{ter}}-\frac{{\left(v_{ter}\right)}^2}{g}\ln 2\\ \approx v_{ter}t+\text{constant}\end{array}$

Conclusion:

Therefore, it is shown that when $t>>\tau$, the position is $y=v_{ter}t+\text{cons}$.

(d)

To determine

Show that when $t$ is small, the position will be $y\approx \frac{1}{2}g{t}^2$.

Answer to Problem 2.35P

It is shown that when $t$ is small, the position will be $y\approx \frac{1}{2}g{t}^2$.

Explanation of Solution

According to Taylor series $\cos hz$ can be written as.

  $\cos hz=1+\frac{1}{2!}{z}^2+\frac{1}{4!}{z}^4+\mathrm{....}$

Substitute, $\frac{gt}{v_{ter}}$ for $z$ in the Taylor series expansion.

  $\cos h\left(\frac{gt}{v_{ter}}\right)=1+\frac{1}{2!}{\left(\frac{gt}{v_{ter}}\right)}^2+\frac{1}{4!}{\left(\frac{gt}{v_{ter}}\right)}^4+\mathrm{....}$

Since $t$ is small, it is possible to neglect $t^4$ term.

  $\cos h\left(\frac{gt}{v_{ter}}\right)\approx 1+{\left(\frac{gt}{v_{ter}}\right)}^2$        (XII)

Take logarithm on both sides of the equation (XII).

  $ln\left[\cos h\left(\frac{gt}{v_{ter}}\right)\right]=\ln \left[1+{\left(\frac{gt}{v_{ter}}\right)}^2\right]$        (XIII)

According to Taylor series expansion.

  $\ln \left(1+\delta \right)=z-\frac{1}{2}{z}^2+\frac{1}{3}{z}^3-\mathrm{...}$

Apply the above Taylor series in equation (XIII).

  $ln\left[\cos h\left(\frac{gt}{v_{ter}}\right)\right]=\frac{g^2{t}^2}{2v_{{}_{ter}}^2}-\frac{1}{2}{\left(\frac{g^2{t}^2}{2v_{{}_{ter}}^2}\right)}^4+\mathrm{..}$        (XIV)

Neglect $t^{}$ terms in equation (XIV).

  $ln\left[\cos h\left(\frac{gt}{v_{ter}}\right)\right]=\frac{g^2{t}^2}{2v_{{}_{ter}}^2}$        (XV)

Substitute, equation (XV) in (IX).

  $\begin{array}{c}y=\frac{{\left(v_{ter}\right)}^2}{g}\left(\frac{g^2{t}^2}{2v_{ter}^2}\right)\\ \approx \frac{1}{2}g{t}^2\end{array}$

Conclusion:

Therefore, It is shown that when $t$ is small, the position will be $y\approx \frac{1}{2}g{t}^2$.