A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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.

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  4. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.