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

  1. May 13, 2007 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. May 13, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    At what angle are you launching the projectile? Are we taking air resistance into account?
  4. May 13, 2007 #3
    The problems given are all based on estimations...there was no angle given...air resistance can be left out, it really doesn't matter either way
  5. May 13, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well, the most effective angle (requiring lowest launch velocity) would be [itex]\pi/4[/itex]. So, tell me, have you any knowledge of kinematic equations?
  6. May 13, 2007 #5
    yes we have touched on them a little bit before
  7. May 15, 2007 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well, we can split the motion into two separate directions; horizontal motion and vertical motion. Can you use kinematic equations to describe the motion of the particle in each of these directions?
  8. May 16, 2007 #7
  9. May 16, 2007 #8
    so the lowest launch velocity, would that be Vi
  10. May 16, 2007 #9
    So vi = pi/4?
  11. May 16, 2007 #10
    V_i is the initial velocity. Pi/4 is the angle that requires the lowest magnitude of V_i. In oter words if theta_i is pi/4, you will get the most distance for any V_i. However, it is up to you to determine what V_i should be. I would personally start by converting the 9 miles into SI Units (I think metric would give some nice numbers to work with.
  12. May 16, 2007 #11
    Yeah, one mile is equivalent to 1.68 km or something like that
    or 1680m

    So, since the horizontal component of velovity is contant

    Range=vt (where v is horizontal component of velocity)

    The vertical motion
    vertical displacement=0

    0=ut -5t^2 (where u is the vertical component of velocity)
  13. May 16, 2007 #12
    9 miles = 14.484096 kilometers
  14. May 16, 2007 #13
    i am going to correct that..it should be 8 miles...
    8 miles = 12.874752 kilometers
  15. May 16, 2007 #14
    yea, so the range= 12.9 km

    on on related note, not enough data is given to solve the equation.
    You need the time,t for the duration of projectile
  16. May 16, 2007 #15
    but what are SI units?
  17. May 16, 2007 #16
    12 874.752 meters
  18. May 16, 2007 #17
    shouldnt be too complicated, i uggest you make it t 3sf
  19. May 16, 2007 #18
    what does t 3sf mean
  20. May 16, 2007 #19
    Okay. Since you said that this problem can be based on estimates, let us call it 12.9 x10^3. (However for 9 miles I am getting 14.5x10^3...) Lokking back at your Projectile motion equations, which one do you think you should use?
  21. May 16, 2007 #20
    i realized it was a typo, im sorry, it should be 8 miles
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Projectile motion