Chapter 2.2, Problem 13E

Interpretation Introduction

Interpretation:

The analytical solution for $\text{v(t)}$ assuming $\text{v(0) = 0}$ and the terminal velocity is to be obtained. Using the graphical analysis of the problem, the formula for terminal velocity is to be re-derived. Also, using the given data in the problem, the average velocity ${\text{(v}}_{\text{avg}}\text{)}$, the terminal velocity, and the drag constant $\text{(k)}$ should be calculated.

Concept Introduction:

The limiting velocity of an object is called its terminal velocity. It is obtained by finding the limit of $\text{v(t) }$ as $t\to \infty$.

The fixed point of the differential equation in terms of $\dot{\text{v}}$ describing the motion of an object represents its terminal velocity.

Answer to Problem 13E

Solution:

a) $\text{v(t) = }\frac{\text{rm}}{\text{k}}\left(\frac{{\text{e}}^{\text{rt}}{\text{-e}}^{\text{rt}}}{{\text{e}}^{\text{rt}}{\text{+e}}^{\text{rt}}}\right)$

b) $\sqrt{\frac{\text{mg}}{\text{k}}}$

c) $\sqrt{\frac{\text{mg}}{\text{k}}}$

d) ${\text{v}}_{\text{avg}}\text{ = 252}\text{.586}\approx \text{ 253 ft/s = 172 mph}$

e) $\text{v}\approx \text{265 ft/s}$ $\text{k =}0.1197$

Explanation of Solution

a) The differential equation which shows the velocity $\text{v(t)}$ of a skydiver falling to the ground is

${\text{m}\dot{\text{v}}\text{ = mg - kv}}^2$

$\begin{array}{l}\text{m}\frac{\text{dv}}{\text{dt}}{\text{ = mg - kv}}^{\text{2}}\\ \text{dt=m}\frac{\text{dv}}{{\text{mg - kv}}^{\text{2}}}\\ \text{dt=}\frac{\text{m}}{\text{k}}\frac{\text{dv}}{{\left(\sqrt{\frac{\text{mg}}{\text{k}}}\right)}^{\text{2}}{\text{- v}}^{\text{2}}}\end{array}$

Integrating with respect to t gives

$\begin{array}{l}\text{t=}\frac{\text{m}}{\text{k}}\sqrt{\frac{\text{k}}{\text{mg}}}{\text{tanh}}^{\text{-1}}\left(\frac{\text{v}}{\sqrt{\frac{\text{mg}}{\text{k}}}}\right)\\ \\ \text{After rearranging we get}\\ \\ \text{v(t) = }\sqrt{\frac{\text{mg}}{\text{k}}}\text{tanh}\sqrt{\frac{\text{gk}}{\text{m}}}\text{t+c}\end{array}$

We are given that $\text{v(0) = 0}$. Hence, the above equation becomes

$\begin{array}{l}\text{0= }\sqrt{\frac{\text{mg}}{\text{k}}}\text{tanh}\sqrt{\frac{\text{gk}}{\text{m}}}\text{(0)+c}\\ \text{c=0}\end{array}$

$\text{v(t) = }\sqrt{\frac{\text{mg}}{\text{k}}}\text{tanh}\sqrt{\frac{\text{gk}}{\text{m}}}t$

$\begin{array}{l}\text{v(t) = }\sqrt{\frac{\text{mg}}{\text{k}}}\frac{{\text{e}}^{\sqrt{\frac{\text{gk}}{\text{m}}}\text{t}}{\text{-e}}^{\sqrt{\frac{\text{gk}}{\text{m}}}\text{t}}}{{\text{e}}^{\sqrt{\frac{\text{gk}}{\text{m}}}\text{t}}{\text{+e}}^{\sqrt{\frac{\text{gk}}{\text{m}}}\text{t}}}\\ \text{v(t) = }\frac{\text{rm}}{\text{k}}\left(\frac{{\text{e}}^{\text{rt}}{\text{-e}}^{\text{rt}}}{{\text{e}}^{\text{rt}}{\text{+e}}^{\text{rt}}}\right)\end{array}$

Where $\text{r = }\sqrt{\frac{\text{gk}}{m}}$

Therefore, the analytical solution for $\text{v(t)}$ is $\text{v(t) = }\frac{\text{rm}}{\text{k}}\left(\frac{{\text{e}}^{\text{rt}}{\text{-e}}^{\text{rt}}}{{\text{e}}^{\text{rt}}{\text{+e}}^{\text{rt}}}\right)$

b) The terminal velocity of the sky diver can be calculated as

$\begin{array}{l}\underset{t\to \infty }{\lim }\text{v(t)}=\sqrt{\frac{\text{mg}}{\text{k}}}\underset{t\to \infty }{\lim }\frac{e^{\sqrt{\frac{\text{gk}}{\text{m}}}t}-e^{\sqrt{\frac{\text{gk}}{\text{m}}}t}}{e^{\sqrt{\frac{\text{gk}}{\text{m}}}t}+e^{\sqrt{\frac{\text{gk}}{\text{m}}}t}}\\ \\ \underset{t\to \infty }{\lim }\text{v(t)}=\sqrt{\frac{\text{mg}}{\text{k}}}\end{array}$

Thus, the terminal velocity of the sky diver is $\sqrt{\frac{\text{mg}}{\text{k}}}$

c) The fixed point of the ${\text{m}\dot{\text{v}}\text{ = mg - kv}}^2$ gives the terminal velocity of the sky diver.

The fixed is obtained by putting $\dot{\text{v}}\text{ = 0}$ in the given differential equation as

$\begin{array}{l}{\text{m(0)= mg - kv}}^{\text{2}}\\ {\text{kv}}^{\text{2}}\text{= mg}\\ \text{ v = }\sqrt{\frac{\text{mg}}{\text{k}}}\end{array}$

Thus, the terminal velocity of the sky diver is $\sqrt{\frac{\text{mg}}{\text{k}}}$

d) The average velocity of the sky diver can be calculated as

$\begin{array}{l}{\text{ v}}_{\text{avg}}\text{=}\frac{{\text{X}}_{\text{f}}{\text{ - X}}_{\text{i}}}{\text{t}}\\ {\text{ v}}_{\text{avg}}\text{=}\frac{\text{31400 - 2100}}{116}\\ {\text{v}}_{\text{avg}}\text{ = 252}\text{.586}\approx \text{ 253 ft/s = 172 mph}\end{array}$

e) We can find the distance fallen s from the equation for v(t) as

$\begin{array}{l}\text{s = }{\displaystyle \int \text{v(t) dt}}\text{ = }{\displaystyle \int \sqrt{\frac{\text{mg}}{\text{k}}}\text{tanh}\sqrt{\frac{\text{gk}}{\text{m}}}\text{t}}\text{ dt}\\ \\ \text{s = }{\displaystyle \int \sqrt{\frac{\text{mg}}{\text{k}}}\frac{\text{sinh}\sqrt{\frac{\text{gk}}{\text{m}}}\text{t}}{\text{cosh}\sqrt{\frac{\text{gk}}{\text{m}}}\text{t}}}\text{ dt}\mathrm{...................}\text{ (1)}\\ \text{Let's substitute u = cosh}\sqrt{\frac{\text{gk}}{\text{m}}}\text{t}\\ \\ \text{After differentiating we get}\\ \\ \text{dt = }\frac{du}{\sqrt{\frac{\text{gk}}{\text{m}}}\text{sinh}\sqrt{\frac{\text{gk}}{\text{m}}}\text{t}}\end{array}$

After substitution equation (1) becomes

$\begin{array}{l}\text{s = }\frac{\text{m}}{\text{k}}{\displaystyle \int \frac{\text{1}}{\text{u}}}\text{ du}\\ \\ \text{s = }\frac{\text{m}}{\text{k}}\text{lnu + C}\\ \text{Substituting back u in terms of t we get}\\ \\ \text{s = }\frac{\text{m}}{\text{k}}\text{ln}\left(\text{cosh}\left(\sqrt{\frac{\text{gk}}{\text{m}}}\text{t}\right)\right)\text{+C}\end{array}$

But $\text{s(0) = 0}$

$\begin{array}{l}\text{Hence,}\\ \text{0 = }\frac{\text{m}}{\text{k}}\text{ln}\left(\text{cosh}\sqrt{\frac{\text{gk}}{\text{m}}}\text{(0)}\right)\text{+C}\\ \text{Thus, the equation for s becomes}\\ \text{s(t) = }\frac{\text{m}}{\text{k}}\text{ln}\left(\text{cosh}\sqrt{\frac{\text{gk}}{\text{m}}}\text{t}\right)\end{array}$

But terminal velocity is $\text{ v =}\sqrt{\frac{\text{mg}}{\text{k}}}$

Therefore,

$\begin{array}{l}\\ \text{s(t) = }\frac{{\text{v}}^{\text{2}}}{\text{g}}\text{ln}\left(\text{cosh}\left(\frac{\text{g}}{\text{v}}\text{t}\right)\right)\\ \end{array}$

After substituting the given values we can write

$\begin{array}{l}\\ \text{29300 = }\frac{{\text{v}}^{\text{2}}}{\text{32}\text{.2}}\text{ln}\left(\text{cosh}\left(\frac{\text{32}\text{.2}}{\text{v}}\text{(116)}\right)\right)\\ \end{array}$

After solving it we get

$\text{v}\approx \text{265 ft/s}$

The drag constant k can be calculated as

$\begin{array}{l}\text{ v = }\sqrt{\frac{\text{mg}}{\text{k}}}\\ \text{ k = }\frac{\text{mg}}{{\text{v}}^{\text{2}}}\\ \text{k = }\frac{(261.2)(32.2)}{{265}^{\text{2}}}\\ \text{k =}0.1197\end{array}$