Forest UCM PnCP ProjMotion

Projectile Motion

Friction depends linearly on velocity

Projectile motion describes the path a mass moving in two dimensions. An example of which is the motion of a projectile shot out of a cannon with an initial velocity $v_0$ with an angle of inclination $\theta$.

When the motion in each dimension is independent, the kinematics are separable giving you two equations of motion that depend on the same time.

Using our solutions for the horizontal and vertical motion when friction depends linearly on velocity (Forest_UCM_PnCP_LinAirRes) we can write :

$x= \frac{m}{b} v_{ix} \left ( 1-e^{-\frac{b}{m}t} \right )$

where $v_{ix}$ has replaced $v_i$ so the components are more explicitly identifiable.

in the y-direction however, the directions are changed to represent an object moving upwards instead of falling

Newton's second law for falling

$\sum \vec{F}_{ext} = mg -bv = m \frac{dv}{dt}$

becomes

$\sum \vec{F}_{ext} = -mg +bv = m \frac{dv}{dt}$

for a rising projectile

This changes the signs in front of the $v_t$ terms such that

$y= v_t t + \frac{m}{b}\left ( v_0 - v_t \right ) \left ( 1- e^{-\frac{b}{m}t} \right )$

becomes

$y = -v_t t + \frac{m}{b}\left ( v_{iy} + v_t \right ) \left ( 1- e^{-\frac{b}{m}t} \right )$

where $v_{iy}$ has replaced $v_0$ so the components are more explicitly identifiable.

We now have a system governed by the following system of two equations

let

$\tau \equiv \frac{m}{b}$

$x= \tau v_{ix} \left ( 1-e^{-\frac{t}{\tau}} \right )$

$y = \tau\left ( v_{iy} + v_t \right ) \left ( 1- e^{-\frac{t}{\tau}} \right ) -v_t t$

Range equation

To determine how far the projectile will travel in the x-direction (Range) you can solve the above equation for $y$ in the case that $y=0$.

since time is the same in both equations you can solve for time in terms of x and substitute for time inthe y-direction equations.

solving for $e^{-\frac{t}{\tau}}$ using the x-direction equation

$x= \tau v_{ix} \left ( 1-e^{-\frac{t}{\tau}} \right )$

$\Rightarrow e^{-\frac{t}{\tau}} = 1- \frac{x }{v_{ix} \tau}$

substituting for $e^{-\frac{t}{\tau}}$

$y = \tau \left ( v_{iy} + v_t \right ) \left ( \frac{x }{v_i \tau} \right ) -v_t t$

$= \frac{ v_{iy} + v_t }{v_i} x -v_t t$

now we need to substitute for time $t$

$e^{-\frac{t}{\tau}} = 1- \frac{x }{v_{ix} \tau}$

$\Rightarrow \ln \left ( e^{-\frac{t}{\tau}} \right) = \ln \left ( 1- \frac{x }{v_{ix} \tau}\right)$

$-\frac{t}{\tau} = \ln \left ( 1- \frac{x }{v_{ix} \tau}\right)$

$t = -\tau\ln \left ( 1- \frac{x }{v_{ix} \tau}\right)$

substituting for time

$y =\frac{ v_{iy} + v_t }{v_{ix}} x -v_t t$

$=\frac{ v_{iy} + v_t }{v_{ix}} x + v_t \tau\ln \left ( 1- \frac{x }{v_{ix} \tau}\right)$

The Range $(R)$ is defined as the value for $x$ when $y =0$

$0 = \frac{ v_{iy} + v_t }{v_{ix}} R + v_t \tau\ln \left ( 1- \frac{R }{v_{ix} \tau}\right)$

$v_{ix} = v_0 \cos \theta$

$v_{iy} = v_0 \sin \theta$

The above equation does not have an exact analytical solution.

You can try to solve it graphically or by taylor expanding small quantities when they appear as arguments to functions like the $\ln$ function

Solution by Taylor expansion

If

ln(1-x) Taylor expansion

Taylor expanding about x = 0 for

$\frac{1}{1+x} = f(x=0) + \left . \frac{d f(x)}{dx} \right |_{x=0} \frac{x-0}{1!} + \left . \frac{d^2 f(x)}{dx^2} \right |_{x=0} \frac{(x-0)^2}{2!} + \cdots = \sum_{n=0}^\infty \left . \frac{d^n f(x)}{dx^n} \right |_{x=0} \frac{(x-0)^n}{n!}$

$=1 - \left . (1+x)^{-2} \right |_{x=0} \frac{(x-0)^1}{1!} + 2 \left . (1+x)^{-3} \right |_{x=0} \frac{(x-0)^2}{2!} + \cdots$

$= 1 -x + 2\frac{(x)^2}{2!} - 2*3 \frac{(x)^3}{3!} + \cdots$

$= 1 -x + x^2 - x^3 + \cdots = \sum_{n=0}^{\infty} (-x)^{n-1}$

$\ln(1+x) = \int \frac{1}{1+x} dx = \int \left ( 1 -x + x^2 - x^3 + \cdots \right ) dx$

$= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$

similarly

$\int \frac{1}{1-x} dx = -\int \frac{1}{u} du \;\;\;\; u = 1-x \;\;\;\; du = -dx$

$\int \frac{1}{1-x} dx = -\ln(u) = -\ln(1-x)$

Taylor expanding about x=0

$\frac{1}{1-x} = f(x=0) + \left . \frac{d f(x)}{dx} \right |_{x=0} \frac{x-0}{1!} + \left . \frac{d^2 f(x)}{dx^2} \right |_{x=0} \frac{(x-0)^2}{2!} + \cdots = \sum_{n=0}^\infty \left . \frac{d^n f(x)}{dx^n} \right |_{x=0} \frac{(x-0)^n}{n!}$

$=1 + \left . (1-x)^{-2} \right |_{x=0} \frac{(x-0)^1}{1!} + 2 \left . (1+x)^{-3} \right |_{x=0} \frac{(x-0)^2}{2!} + \cdots$

$= 1 + x + 2\frac{(x)^2}{2!} + 2*3 \frac{(x)^3}{3!} + \cdots$

$= 1 +x + x^2 +x^3 + \cdots = \sum_{n=0}^{\infty} (x)^{n-1}$

$\int \frac{1}{1-x} dx = -\ln(1-x) = \int \left ( 1 +x + x^2 +x^3 + \cdots \right ) dx$

$-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots$

$\ln(1-x) = - \left ( x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots \right )$

Taylor expand ln term in Range equation

$\ln \left ( 1- \frac{R }{v_{ix} \tau}\right) = - \left ( \frac{R }{v_{ix} \tau} + \frac{\left ( \frac{R }{v_{ix} \tau} \right ) ^2}{2}+ \frac{\left ( \frac{R }{v_{ix} \tau} \right ) ^3}{3} + \cdots \right )$

$\frac{ v_{iy} + v_t }{v_{ix}} R = v_t \tau \left ( \frac{R }{v_{ix} \tau} + \frac{\left ( \frac{R }{v_{ix} \tau} \right ) ^2}{2}+ \frac{\left ( \frac{R }{v_{ix} \tau} \right ) ^3}{3} + \cdots \right )$

$R=0$ is a solution $\Rightarrow$ height is zero when $R = 0$

$\frac{ v_{iy} + v_t }{v_{ix}} = v_t \tau \left ( \frac{1 }{v_{ix} \tau} + \frac{R }{2v_{ix}^2 \tau^2}+ \frac{R^2 }{3 v_{ix}^3 \tau^3} + \cdots \right )$

$v_{iy} + v_t = v_t \left ( 1 + \frac{R }{2v_{ix} \tau}+ \frac{R^2 }{3 v_{ix}^2 \tau^2} + \cdots \right )$

$v_{iy} = v_t\left ( \frac{R }{2v_{ix} \tau}+ \frac{R^2 }{3 v_{ix}^2 \tau^2} + \cdots \right )$

$\frac{R }{2v_{ix} \tau} \sim \frac{v_{iy}}{v_t} - \frac{R^2 }{3 v_{ix}^2 \tau^2}$

$R \sim \frac{2v_{iy}v_{ix} \tau}{v_t} - \frac{2 R^2 v_{ix} \tau}{3 v_{ix}^2 \tau^2}$

$\sim \frac{2v_{iy}v_{ix}}{g} - \frac{2 }{3 v_{ix} \tau} R^2$

$\sim R_{vacuum} - \frac{2 }{3 v_{ix} \tau} R^2$

Forest_UCM_PnCP#Projecile_Motion