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!

Solving for T in a horizontal projectile equation

  1. Apr 12, 2003 #1
    Hi, I am stuck on even how to start to solve for T

    the equations is:

    0 = Vi / K (1 - e^-kt)(sin theta) + (g / k^2)(1-kt-e^-kt)

    Any suggestions on how to begin to solve for T would be appreciated.

    Thanks,

    Matt
     
  2. jcsd
  3. Apr 12, 2003 #2

    FZ+

    User Avatar

    My oh my is this complicated....

    Hmm... your problem is this bit:

    1-kt-e^-kt

    Since you have t both inside and outside the e^, I don't think there is an algebraic method for you to get an exact answer. Are you sure you formed the equation correctly?

    I may be wrong though...
     
  4. Apr 12, 2003 #3
    Yes, it is correct :(
     
  5. Apr 13, 2003 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    HallsofIvy

    Then you will need either to use a numerical form of solution or do a google search form "Lambert W function".
     
  6. Apr 13, 2003 #5

    enigma

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Couldn't you do it iteratively?

    Pick a value for T and solve. Then take that result and plug it in for T and repeat.

    If all goes well (depending on your pick to start...), it will converge on an answer.
     
  7. Apr 14, 2003 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Pick a value of T and solve for what? :smile: What you are describing (I think!) is one very crude way of solving an equation iteratively.

    Newton's method will work faster.
     
  8. Apr 14, 2003 #7
    Sorry if this seems dumb ...
    But how did the equation of an horizontal projectil get this complicated ?
    The horizontal projectile equations that i know are well too easier ! (The one derived from the SUVAT equations)
     
  9. Apr 14, 2003 #8
    The equation is for air drag, notice the k.
     
  10. Apr 14, 2003 #9
    Ok, I got it down to

    10^1+.5t = 19.6t

    Can anyone help me on the Omega function?
     
  11. Apr 15, 2003 #10
    This is what I did:

    I inserted some fixed constants and multiplied out

    (48/.5)(1-^e-.5t)(sin 45) + (9.8/.25)(1-(.5t) - e^-.5t) = 0

    (96 - 96e^-.5t)(sin 45) + 39.2(1-(.5t)-e^-.5t) = 0

    67.88 - 67.88^e-.5t + 39.2 - 19.6t - 19.6e^-.5 = 0

    107.2 - 87.48e^-.5t - 19.6t = 0

    107.2 - 19.6t = 87.48^e-.5t

    log(107.2 - 19.6t) = log(87.48^e-.5t)

    log107.2 - log19.6t = -.5tLog(87.48)

    2.030194 - log19.6t = -.5t(1.94198)

    1.045463 - log19.6t = -.5t

    -(1.045463 - log19.6t = -.5t)

    -1.045463 + log19.6t = .5t

    log19.6t = .5t + 1.045463

    10^(.5t + 1.045463) = 19.6t

    This is where Im stuck.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Solving for T in a horizontal projectile equation
  1. Equation solving (Replies: 5)

  2. Solve for T (Replies: 2)

  3. Solving equation (Replies: 4)

  4. Solve an equation (Replies: 8)

Loading...