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 [math]v_0[/math] with an angle of inclination [math]\theta[/math].
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 :
- [math]x= \frac{m}{b} v_i \left ( 1-e^{-\frac{b}{m}t} \right )[/math]
in the y-direction however, the directions are changed to represent an object moving upwards instead of falling
Newton's second law for falling
- [math]\sum \vec{F}_{ext} = mg -bv = m \frac{dv}{dt}[/math]
becomes
- [math]\sum \vec{F}_{ext} = -mg +bv = m \frac{dv}{dt}[/math]
for a rising projectile
This changes the signs in front of the [math]v_t[/math] terms such that
- [math]y= v_t t + \frac{m}{b}\left ( v_0 - v_t \right ) \left ( 1- e^{-\frac{b}{m}t} \right ) [/math]
becomes
- [math]y = -v_t t + \frac{m}{b}\left ( v_0 + v_t \right ) \left ( 1- e^{-\frac{b}{m}t} \right ) [/math]
We now have a system governed by the following system of two equations
let
[math]\tau \equiv \frac{m}{b}[/math]
- [math]x= \tau v_i \left ( 1-e^{-\frac{t}{\tau}} \right )[/math]
- [math]y = \tau\left ( v_0 + v_t \right ) \left ( 1- e^{-\frac{t}{\tau}} \right ) -v_t t[/math]
Range equation
To determine how far the projectile will travel in the x-direction (Range) you can solve the above equation for [math]y[/math] in the case that [math]y=0[/math].
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 [math]e^{-\frac{t}{\tau}}[/math] using the x-direction equation
- [math]x= \tau v_i \left ( 1-e^{-\frac{t}{\tau}} \right )[/math]
- [math]\Rightarrow e^{-\frac{t}{\tau}} = 1- \frac{x }{v_i \tau}[/math]
substituting for [math]e^{-\frac{t}{\tau}}[/math]
- [math]y = \tau \left ( v_0 + v_t \right ) \left ( \frac{x }{v_i \tau} \right ) -v_t t[/math]
- [math]= \frac{ v_0 + v_t }{v_i} x -v_t t[/math]
now we need to substitute for time [math]t[/math]
- [math] e^{-\frac{t}{\tau}} = 1- \frac{x }{v_i \tau}[/math]
- [math]\Rightarrow \ln \left ( e^{-\frac{t}{\tau}} \right) = \ln \left ( 1- \frac{x }{v_i \tau}\right)[/math]
- [math]-\frac{t}{\tau} = \ln \left ( 1- \frac{x }{v_i \tau}\right)[/math]
- [math]t = -\tau\ln \left ( 1- \frac{x }{v_i \tau}\right)[/math]
substituting for time
- [math]y =\frac{ v_0 + v_t }{v_i} x -v_t t[/math]
- [math] =\frac{ v_0 + v_t }{v_i} x + v_t \tau\ln \left ( 1- \frac{x }{v_i \tau}\right)[/math]
The Range [math](R)[/math] is defined as the value for [math]x[/math] when [math]y =0[/math]
- [math]0 = \frac{ v_0 + v_t }{v_i} R + v_t \tau\ln \left ( 1- \frac{R }{v_i \tau}\right)[/math]
Forest_UCM_PnCP#Projecile_Motion