bishnu

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)

how would i calculate the velocity coming out of an air cannon in terms of pressure, volume,barrel length, area of barrel opening, k(coffecient of porpotionalityin -kv^2 for friction)?

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

i know it is wrong becasue it doesnt output corrcect numbers

I would but i want to take in to account that the force acting on the projectile isnt constant and i want my equation to use pressure and i also want friction

Im not posting this in the physics forum because i feel its more of a amth question. My friends and i have bulit an air cannon and i wish to analysis it mathematically...after a couple of hours i came up with an equation i think would do. I was wondering if people could look at my work and see of they see anything wrong with it. First off my aircannon has a simple chamber then a valve and then a barrel. You pump air in to the camber and then let relase the valve and then the projectile is shot(ps i have never used LaTex before so we will see how this goes) first off i started with the equation [math] p=\frac{F}{A}[/math] In this problem i am going to assume the barrel fits perfectly in the barrel so there is no room inbetween the barrel and projectile. [math]pA_b=F[/math] [math] p_iv_i=p_fv_f[/math] [math]\frac{p_i*v_i}{v^i+A_b*x}[/math]ab= area of the barrel and x=distance traveled in the barrel [math]f=\frac{p_iv_iaf_b}{v_i+a_bx}[/math] this equation tells the force on the projectile at anytime druring the barrel friction=kv^2 k=constant of proptionality [math]f=\frac{p_iv_ia_b}{v_i+a_bx}-kv^2[/math] now intergrate both sides from 0 to the length of the barrel xb to get the energy now im getting lazy of typing in all that stuff so ill just tell you what i did from here on i substituted 1/2mv^2 for energy now i just sepertated the v and keep the v on the other side in the friction equation. Now i just subsitituted this eqaution into the v in the friction part and did that infintly many times which gives you 2 infinite series they converege and i did a little algebra to reduce it and i got [math]v=(\frac{2}{m_p+2kx_b}(p_iv_i\ln({\frac{a_bx_b}{v_i}+1})+\frac{2k^2x^2_b}{m_p}))^(1/2)[/math] i want to know if anyone can spot any errors

Okay this problem is relativity easy...first find the Maclaurin Series for arctan and since it is going to 0 you can use the series in the interaval [-1,1] the maculaurin series is x-x^3/3+x^5/5-x^7/7...now subsititue the series in for arctan u get x( 1/x-1/3x^3+1/x5^5...) distrubte 1-1/3x^2+1/5x^4... and then plug in zero for the limit and you get undefined...which means it oscillates like arcsin and arccos. A way to think about it is when ever you have 1/x it means that you basically flip the coordinate grid all the values that go to zero instead goto infinity and vice versa so the question is basically asking what will arctan be at infinity which is unknowable or undefined.