Difference between revisions of "Forest UCM PnCP ProjMotion"

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This changes the signs in front of the <math>v_t</math> terms such that  
 
This changes the signs in front of the <math>v_t</math> terms such that  
  
:<math>y= v_t t + \frac{1}{b}\left ( v_0 - v_t) \right ) \left ( 1- e^{-bt}  \right )  </math>
+
:<math>y= v_t t + \frac{1}{b}\left ( v_0 - v_t \right ) \left ( 1- e^{-bt}  \right )  </math>
  
 
becomes
 
becomes
  
:<math>y = -v_t t + \frac{1}{b}\left ( v_0 + v_t) \right ) \left ( 1- e^{-bt}  \right ) </math>
+
:<math>y = -v_t t + \frac{1}{b}\left ( v_0 + v_t \right ) \left ( 1- e^{-bt}  \right ) </math>
  
 
We now have a system governed by the following system of two equations
 
We now have a system governed by the following system of two equations
  
:<math>y =  \frac{1}{b}\left ( v_0 + v_t) \right ) \left ( 1- e^{-bt}  \right )  -v_t t</math>
+
let
 +
 
 +
<math>\tau \equiv \frac{m}{b}</math>
 +
 
 +
:<math>x=  \tau v_i \left ( 1-e^{-\frac{b}{m}t} \right )</math>
 +
:<math>y =  \frac{1}{b}\left ( v_0 + v_t \right ) \left ( 1- e^{-bt}  \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^{-bt}</math> using the x-direction equation
 +
 
 +
:<math>x=  \frac{m}{b} v_i \left ( 1-e^{-\frac{b}{m}t} \right )</math>
 +
:\Rightarrow e^{-bt} =\frac{x b}{m v_i} \left ( 1
 +
 
 +
:<math>y =  \frac{1}{b}\left ( v_0 + v_t \right ) \left ( 1- e^{-bt}  \right )  -v_t t</math>
  
  
 
[[Forest_UCM_PnCP#Projecile_Motion]]
 
[[Forest_UCM_PnCP#Projecile_Motion]]

Revision as of 12:35, 1 September 2014

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{1}{b}\left ( v_0 - v_t \right ) \left ( 1- e^{-bt} \right ) [/math]

becomes

[math]y = -v_t t + \frac{1}{b}\left ( v_0 + v_t \right ) \left ( 1- e^{-bt} \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{b}{m}t} \right )[/math]
[math]y = \frac{1}{b}\left ( v_0 + v_t \right ) \left ( 1- e^{-bt} \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^{-bt}[/math] using the x-direction equation

[math]x= \frac{m}{b} v_i \left ( 1-e^{-\frac{b}{m}t} \right )[/math]
\Rightarrow e^{-bt} =\frac{x b}{m v_i} \left ( 1
[math]y = \frac{1}{b}\left ( v_0 + v_t \right ) \left ( 1- e^{-bt} \right ) -v_t t[/math]


Forest_UCM_PnCP#Projecile_Motion