1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Projectile motion with friction?

  1. Dec 12, 2008 #1
    If there is air resistance, when projectile motion occurs, in what angle we throw the object can we obtain the highest range?

    Can anybody please help me with this question? I have been doubtful about this question. My teacher says it has something to do with calculus but he has already forgotten some rules of calculas.

    Can anybody answer this question? If you show how you can arrive this question, it will be better. Thank you.
     
  2. jcsd
  3. Dec 12, 2008 #2

    olgranpappy

    User Avatar
    Homework Helper

    It depends on the form of the air resistance contribution to the force.
     
  4. Dec 12, 2008 #3

    olgranpappy

    User Avatar
    Homework Helper

    ...but, it the resistant force is assumed linear in the velocity the I think the problem isn't too hard to work out. Write down the equations of motion and we can work through it together if you want. Cheers.
     
  5. Dec 13, 2008 #4
    I think the answer is that nobody knows, but folks around here seem to disagree with me. I think that the flight of an object through a gas is highly non-trivial. The two prevailing approximations are that the force of resistance due to drag from the gas is proportional to either the first power or the square of the velocity, with the proportionality constant determined from experiment. To truly do this problem I think you would need to solve a compressible Navier-Stokes setup. CrazY!
     
  6. Dec 13, 2008 #5
    About 34 degrees
     
  7. Dec 13, 2008 #6
    Hey, How Phrak works the answer out?

    Olgranpappy, Here is the equation:
    mg+bv=ma for veritcal motion of the thing.

    bv=ma for horizontal motion.

    I sincerely believe that the result requires something to do with calculus, can anybody who is excellent in calculus help me to deal with the problem?
     
  8. Dec 13, 2008 #7

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Okay, let's start with the vertical component. Acceleration is the second derivative of position with respect to time and velocity if the first derivative of position with respect to time. Hence, your equation may be written

    [tex]m\frac{d^2y}{dt^2} = b\frac{dy}{dt} + mg[/tex]

    or in canonical form:

    [tex]\frac{d^2y}{dt^2} - \frac{b}{m}\frac{dy}{dt} = g[/tex]

    Which is a second order linear ODE with constant coefficients. Have you met such equations before?

    As this question is certainly in the style of a homework question, I'm moving it to Into. Phys. in the Homework forums.
     
  9. Dec 13, 2008 #8
    I didn't work it out. I was being flippant. Including air friction, the angle will always be somewhere less than 45 degrees in still air. How many degrees is dependent upon a lot of conditions including Reynolds number.

    @Hootenanny. Viscous drag occurs at very low speeds and is proportional to the velocity. But in most worldly cases you'd be dealing with turbulent drag, proportional to the square of the velocity, as olgranpappy may have been implying.

    [tex] m \ddot {x} = -mg - c \dot{x}^2 [/tex]

    https://www.physicsforums.com/newreply.php?do=newreply&p=816101
     
  10. Dec 13, 2008 #9

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes I know. However, it is entirely up to the OP how they want to define the parameters of the problem. The OP has already been made aware that the result is highly dependent upon the form of the drag, whether quadratic or otherwise.

    The OP was asked to decide on the form of the drag and responded with the linearised form. Therefore, my response was based on the OP's decision to consider linear drag.
     
  11. Dec 13, 2008 #10

    olgranpappy

    User Avatar
    Homework Helper

    It certainly does have something to do with calculus--namely, we have to solve differential equations.

    In order to make our problem well defined I think that the assumption of a linear drag force is fine. In this case we will be able to solve the equations of motion exactly. Unfortunately, we will find that we cannot solve for the range (at least I couldn't) in terms of simple well-known functions.
     
  12. Dec 13, 2008 #11

    olgranpappy

    User Avatar
    Homework Helper

    There may be a sign error in the above equation. I think, +mg should be -mg... assuming that we take "up" as positive... isn't this usually what we do? I guess it's a convention... Also, I usually prefer to define constants such as "b" without any hidden minus signs, so I would probably flip the sign on "b" as well.

    But regardless it will be good to know whether the OP has seen linear differential equations before.
     
  13. Dec 13, 2008 #12

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It's purely a matter of taste for me. I usually choose all my coefficients such that all the signs in the DE are positive, but I concede that it can be confusing, particularly when trying to keep a track of directions. So yes, it would probably be more informative to re-write the ODE

    [tex]\ddot{y}-b\dot{y} = -mg[/tex]

    Where b,m,g are positive constants.
     
  14. Dec 14, 2008 #13
    ODE.............
    I have not learnt it yet.

    Are you two sure that it is less than 45 degree for the max. range? Why you cannot work out the range?What is the final equation you obtain? Is it involved both cos and sin?
     
  15. Dec 14, 2008 #14

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    The range can be worked out, but doing that requires solving an Ordinary Differential Equation (ODE), which you haven't learned about yet.

    The idea of this forum is to help YOU obtain the equation, not to have other people do all the work for you.

    But honestly, I am having trouble seeing the benefit in assigning a differential equations problem to students who are just beginning calculus. This is more a comment about your teacher, and not about you personally.
     
  16. Dec 14, 2008 #15

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I totally agree. In fact, I was shocked to read that
    This is indeed worrying. I'm appalled that someone teaching what appears to be college level physics has forgotten calculus and doesn't appear to know how to solve basic ODE's beyond that it has something to do with calculus.
     
  17. Dec 14, 2008 #16
    Loup, this isn't a homework assignment, is it?
    What class is this for, and what grade are you in?
     
  18. Dec 14, 2008 #17

    olgranpappy

    User Avatar
    Homework Helper

    well... since this isn't really a homework problem maybe it is okay to give a few equations explictly as examples instead of being so very Socratic...

    For example, to find the time of flight (T) you would have to solve the equation:
    [tex]
    0=v_0\sin(\phi)\left(\frac{1-e^{(-\beta T)}}{\beta}\right)-\frac{g}{\beta}\left(T-\frac{1-e^{(-\beta T)}}{\beta}\right)\;,
    [/tex]
    where [itex]\phi[/itex] is the angle that the launcher makes with the horizontal, and where g is 9.8m/s^2 and where [itex]\beta=\alpha/m[/itex], where [itex]\alpha[/itex] determines the resistant force as
    [tex]
    \bold{F}_{\rm res}=-\alpha\bold{v}\;,
    [/tex]
    where v is the velocity and v_0 is the initial velocity.

    The problem is not just the that equation is messy, the problem is that the equation involves T in both a polynomial and an exponential and so cannot be solved for in terms of elementary functions. But, it is possible to do it numerically without too much effort.

    Then, given the time of flight one can plug into a similar equation for the x-position to find the range. And then take a dirivative with respect to \phi and set that equal to zero to find the angle which maximizes the range.

    Cheers.
     
  19. Dec 14, 2008 #18
    Msg me if you want the equation for time to reach termial velocity or distance. Just did it a hour ago, took me 4 days...
     
  20. Dec 15, 2008 #19
    olgranpappy, you are kind. Indeed it is not a homework question. It is simply for my interest. The exams or tests will even not ask this kind of questions. It is just too difficult. I am a Secondary 6 student in Hong Kong. I don't know how to translate that to grade to make you understand :(

    To others, Don't you blame my teacher. He is a nice guy, Since it is not in the syllabus, the teacher hasn't done this kind of Mathematics for ages. It is normal to forget. It is not he forgot how to do basic calculus, but the more deep calculus like ODE.

    In fact, I know about the 2 equations but I don;t know how to solve them. My teacher says it is less than 45 degrees and I doubt. People shouldn't believe something unless proof is given. That's why I ask the question here, in order to find someone who can solve the equations for me. If it is something I can cope with, I will have no need to ask the question here. By the way, I did raise the 2 equations myself!

    Here in Hong Kong, peole who learn Physics have no need to learn about Pure Mathematics. I take the course, Physics, Chemistry and Biology. I am reading a Pure Mathematics books myself in order to cope with the difficutlies I met in Physics. But it requries time and before I am finish the books, I need you guys to help me. Thanks a lot for your help.

    Maybe you will think I am rather rude in this message, am I ? Please forgive me, I am not good at speaking and talking to others.
     
  21. Dec 15, 2008 #20

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    If it's not an assigned homework problem, then that is another matter. No, you're not being rude, and it sounds like you have a good teacher.

    I've been able to verify olgranpappy's method, so yes the solution requires numerical methods. There is no equation using familiar functions that will provide the answer.

    Since terminal velocity is never reached, can you clarify what you really mean?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?