Question

Projectile Motion with Air Resistance Determine a system of differential equations that describes the path of motion in Problem 23 if air resistance is a retarding force $\mathit{k}$(of magnitude $\mathit{k}$) acting tangent to the path of the projectile but opposite to its motion. See Figure 4.9.3. Solve the system. [Hint: $\mathit{k}$is a multiple of velocity, say, role="math" localid="1668407948103" $\mathit{\beta }\mathit{\nu }$.]

Short answer

$$\begin{gathered} x\left(t\right)=c_1+c_2{e}^{-\frac{\beta }{m}t} \\ y\left(t\right)=c_1+c_2{e}^{-\frac{\beta }{m}t}-\frac{mg}{2\beta }t \end{gathered}$$

Step-by-step solution

To Find the differential equation

We have a projectile shot from a gun is being affected by its weight (mg) and a retarding air resistance, and we have to obtain the differential equation 's system that describes its path as the following:

If we chose the right and the upward directions are the positive directions, then according to the figure below by applying newton 's law we can have differential equation's system as

$$\begin{gathered} m\frac{d^{2}x}{d{t}^2}=-k\cos \theta \\ m\frac{d^{2}y}{d{t}^2}=-mg-k\sin \theta \end{gathered}$$

which is

$$\begin{gathered} m\frac{d^{2}x}{d{t}^2}=-\beta v\cos \theta \\ m\frac{d^{2}y}{d{t}^2}=-mg-\beta v\sin \theta \end{gathered}$$

Since we have the horizontal and vertical components as$\frac{dx}{dt}=v\cos \theta$and $\frac{dy}{dt}=v\sin \theta$, then we can have

$$\begin{gathered} \cos \theta =\frac{1}{v}\frac{dx}{dt} \\ \sin \theta =\frac{1}{v}\frac{dy}{dt} \end{gathered}$$

After that, substitute from equation (c) into equation (a) and from equation (d) into equation (b), then we have

$$\begin{gathered} m\frac{d^{2}x}{d{t}^2}=-\beta v\left(\frac{1}{v}\frac{dx}{dt}\right) \\ m\frac{d^{2}y}{d{t}^2}=-mg-\beta v\left(\frac{1}{v}\frac{dy}{dt}\right) \end{gathered}$$

Then we have

localid="1668409441153" $$\begin{gathered} \frac{d^{2}x}{d{t}^2}+\frac{\beta }{m}\frac{dx}{dt}=0 \\ \frac{d^{2}y}{d{t}^2}+\frac{\beta }{m}\frac{dy}{dt}=-g \end{gathered}$$

For the first differential equation : assume that $x=e^{rt}$, then differentiate with respect to t, then we have

$$\begin{gathered} \frac{dx}{dt}=r{e}^{rt} \\ \frac{d^{2}x}{d{t}^2}=r^2{e}^{rt} \end{gathered}$$

After that, substitute from equations (1a) and (1b) into equation (1), then we have

$$\begin{gathered} r^2{e}^{rt}+\frac{\beta }{m}r{e}^{rt}=0 \\ \left(r^2+\frac{\beta }{m}r\right)e^{rt}=0 \end{gathered}$$

Since $e^{rt}$can not be equal 0, then we can have the auxiliary equation of our D.E as

$$\begin{gathered} r^2+\frac{\beta }{m}r=0 \\ r\left(r+\frac{\beta }{m}\right)=0 \end{gathered}$$

Then we have the roots as

$r_1=0\text{and}{r}_2=-\frac{\beta }{m}$

Then we can obtain the general solution of the first differential equation (1) as

$x\left(t\right)=c_1+c_2{e}^{-\frac{\beta }{m}t}$

The second differential equation

For the second differential equation shown in equation (2), we can obtain its solution as the following:

First, we have to obtain the solution of the corresponding homogeneous D.E

$\frac{d^{2}y}{d{t}^2}+\frac{\beta }{m}\frac{dy}{dt}=0$

by assuming that $y=e^{rt}$, then differentiate with respect to t, then we have

$$\begin{gathered} \frac{dy}{dt}=r{e}^{rt} \\ \frac{d^{2}y}{d{t}^2}=r^2{e}^{rt} \end{gathered}$$

After that, substitute from equations (2a) and (2b) into equation (3), then we have

$$\begin{gathered} r^2{e}^{rt}+\frac{\beta }{m}r{e}^{rt}=0 \\ \left(r^2+\frac{\beta }{m}r\right)e^{rt}=0 \end{gathered}$$

Since can not be equal 0 , then we can have the auxiliary equation of our D.E as

$$\begin{gathered} r^2+\frac{\beta }{m}r=0 \\ r\left(r+\frac{\beta }{m}\right)=0 \end{gathered}$$

Then we have the roots as

$r_1=0\text{and}{r}_2=-\frac{\beta }{m}$

Then we can obtain the general solution of the first differential equation (1) as After that, substitute from equations (2a) and (2b) into equation (3), then we have

$$\begin{gathered} r^2{e}^{rt}+\frac{\beta }{m}r{e}^{rt}=0 \\ \left(r^2+\frac{\beta }{m}r\right)e^{rt}=0 \end{gathered}$$

Since$e^{rt}$can not be equal 0 , then we can have the auxiliary equation of our D.E as

$$\begin{gathered} r^2+\frac{\beta }{m}r=0 \\ r\left(r+\frac{\beta }{m}\right)=0 \end{gathered}$$

Then we have the roots as

$r_1=0\text{and}{r}_2=-\frac{\beta }{m}$

Then we can obtain the general solution of the first differential equation (1) as

$y_c\left(t\right)=c_1+c_2{e}^{-\frac{\beta }{m}t}$

Second, we have to obtain the particular solution by assuming that $y_p=At$, then differentiate with respect to as

$$\begin{gathered} \frac{d{y}_p}{dt}=2A \\ \frac{d^2{y}_p}{d{t}^2}=0 \end{gathered}$$

After that, substitute from equations (3a) and (3b) into equation (2), then we have

$0+\frac{2\beta }{m}A=-g$

then we have : $A=-\frac{mg}{2\beta }$

Then we have the particular solution as

$y_p=-\frac{mg}{2\beta }t$

Then from equations (4) and (5), we can obtain the solution of the second differential equation (2) as

$$\begin{gathered} y\left(t\right)=y_c+y_p \\ =c_1+c_2{e}^{-\frac{\beta }{m}t}-\frac{mg}{2\beta }t \end{gathered}$$

Final Answer

$$\begin{gathered} x\left(t\right)=c_1+c_2{e}^{-\frac{\beta }{m}t} \\ y\left(t\right)=c_1+c_2{e}^{-\frac{\beta }{m}t}-\frac{mg}{2\beta }t \end{gathered}$$