I am a senior in high school (second day), with a background in ap physics and precalculus (i do know some calc though). I have been trying to derive kinematic equations with a dissaptive force involved(ie friction). I have tried a couple of methods...i want to know if i got the right answer

first method:

first off i assume that the friction is porpotional to the speed(i hear expermentaly it is more -kv^2)

[math] y''=a-\tfrac{-ky'}{m_p} [/math]

[math]\tfrac{d^2x}{d^2t}+\tfrac{kdx}{m_pdt}-a=0[/math]

find the roots

[math]x=c_1e^(t(\tfrac{-k}{2m_p}+\sqrt{\tfrac{k^2}{4m_p^2}+a}))+c_2e^(t ( \tfrac{-k}{2m_p}-\sqrt{\tfrac{k^2}{4m_p^2}+a}))[/math]

then take the dervative of that to find velocity

second method:

with -kv^2

every moments energy can be expressed in(the subset i represents the moment before hand)

[math]E_(total)=.5mv_i^2+mgx-kxv_i^2 [/math]

substitute mv^2/2 for etotal and do some algebra andsince it is a moment divide each x by n where n is defined as the amount of itertations on the function

[math]v=\sqrt{v_i^2(1-\frac{2kx}{mn})+\frac{2gx}{n}}[/math]

now if you write an equation for the amount of iterations you do

by pluging v into v_i

[math]v=\sqrt{v_i^2(1-\frac{2kx}{mn})^n+\frac{2gx}{n} \sum_{i=0}^{n} (1-\frac{2k}{mn})^i[/math]

set n to infintiy and get

[math]v=\sqrt{v_i^2e^(-2kx)+\frac{gm}{k}}[/math]

i want to know what that means is it the limit of the function or just garabage how could i change my method to an answer that isnt a limit

It would be nice if you could explain what you are doing in a little more detail (particularly your first two equations). Anywho...I would approach this problem by first using Netwon's 2. law on the projectile, i.e.

 

ma(t) = mg + f(t)

 

where f(t) is the force of friction. Note that I've written certain terms in bold to indicate they are vectors. You'll have to remember that f(t) not only changes in magnitude but also in direction with time. In the end, you're going to have to solve a pair of diff. equations (which you may or may not know how to solve, given your background).

okay sorry i didnt explain better

 

y=distance

y'=velocity

y''=acceleration

a=constant acceleration

y''=a this is one of the common newton kinematic equation

-frictional acceleration which is -kv/mp

y''=a-ky'/mp this is what i belive the equation would be for a damped kinematic equation would be

 

the second equation was just me restaing the first but with the derivatives and then i justsolved the differntal equation

 

this is just the direction of the projectile in the y-axes (because it experences acceleration)

okay sorry i didnt explain better

a=constant acceleration

I'll assume that by a you actually mean g (acceleration due to gravity).

 

y''=a this is one of the common newton kinematic equation

So you're saying the projectile's acceleration is constant?

 

y''=a-ky'/mp this is what i belive the equation would be for a damped kinematic equation would be

Why would you believe this? Physics doesn't work that way.

 

this is just the direction of the projectile in the y-axes (because it experences acceleration)

The projectile also experiences acceleration in the horizontal direction (due to friction), so if you want a complete description of the projectile's motion, you'll have to deal with this also.

The acceleration in the equation is constant becuase it is only experencing gravity(im considering the changes in gravity due to height to be trival.

 

I dont understand by what you mean physics doesnt work that way

 

well you offered a solution using vector calaculus instead im just going to right an equation for the forces experenced in the y and the x axis and find the roots on the y and take the time for that and plug it into the x equation

I dont understand by what you mean physics doesnt work that way

I said this because of the explanation you gave me for your equation: " y''=a-ky'/mp this is what i belive...". Don't believe, derive and explain.

 

well you offered a solution using vector calaculus instead im just going to right an equation for the forces experenced in the y and the x axis and find the roots on the y and take the time for that and plug it into the x equation

I didn't use any vector calculus. The equation I gave you is in vector form, but you can easily break it up into components and solve for each individually.

i put in the diff forum because im terible at solving diff equations so i was hoping someone would verify the formula i derived in my first post and the meaning of the constants

I tried solving using a different method from yours by writing vectorial differential equations for acceleartion, velocity and position. It becomes easier, you can calculate position as a function of time and velocity as a function of time. Try it out, if you are unable to do it, i'll post up my complete solution.