## anonymous 5 years ago A gun is fired straight up. assuming that the air drag on the bullet caries quadratically with speed, show that the speed varies with the height according to the equations: v^2= A*e^(2kx) -g/k (upward motoin) and v^2= g/k - B*e^(2kx) (downward motion) A and B are constants of integraion, and k=C2/m, where C2 is the drag constant, and m is the mass of the bullet.

1. anonymous

btw, I will give GOLD! This is what I have done so far: so I can find V(naught) as a function of X (position), there are my assumtions; Up is positive, V(final)=0, and Xinitial=0. I will call C2(drag constant) : c for simplicity. ma=mg+cv^2 mdv/dt= mg+cv^2 mvdv/dx= mg+cv^2 (where dv/dt= (dv/dx)*(dx/dt)) and now: mvdv/dx=mg(1+cv^2/mg) vdv/dx=g(1+cv^2/mg) (v/(1+cv^2/mg)) dv= gdx after simplifying I get: x= m/2c ln(1+cv^2/mg) Now I don't know how to find x as a function of v^2 like in the suggested equations in the question.

2. JamesJ

This is hard to read, but if your last equation is correct: $x= m/2c . \ln(1+cv^2/mg)$ then $\ln(1 + cv^2/mg) = 2cx/m$ $cv^2/mg + 1 = e^{2cx/m}$ $v^2 = \frac{mg}{c} (e^{2cx/m} - 1)$

3. anonymous

yeh but it does not match the answer they give for upward motion. I did it again now and I got this: -mg/2c ln(1-cv^2/mg)=x and when I solve for v^2 $-2cx/mg= \ln(1-cv^2/mg)$ $-1+e^-2cx/mg =cv^2/mg$ $-mg/c+(mg/c)*e^{-2c/mg} = v^2$ now, as told in the question k=c/m, then: $Ae^{-2k}-g/k = v^2$ which is the same expression as the question where A=g/k I think I got how to do the rest of the question, Thanks James