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!

Kinematics in Two Dimensions?

  1. Sep 4, 2004 #1
    Kinematics in Two Dimensions??

    Hi, I was wondering if anyone could give a little help on this problem. I can't figure out where to begin!
    An Olympic long jumper leaves the ground at an angle of 23 degrees and travels through the air for a horizontal distance of 8.7m before landing. What is the take off speed of the jumper?
    Thanks!!
     
  2. jcsd
  3. Sep 4, 2004 #2

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    Let x be horizontal and y be vertical and suppose the jumper takes off from the origin (0, 0) at t = 0. Then
    [tex]x = v_0 \cos( \theta) t[/tex]
    and
    [tex]y = v_0 \sin(\theta) t - \frac{1}{2}g t^2[/tex]
    so, first, find the time at which he/she lands (y = 0) and use that value of t in the first equation to find [itex]v_0[/itex]
     
  4. Sep 4, 2004 #3
    Why is y=0 and for the y equation given is V=0?
     
  5. Sep 4, 2004 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    y=0 on landing because he is landing on the ground which where you are measuring from!

    No, v0 is the same in both equations. (It is the sine and cosine that distinguish vertical and horizontal speeds.) That means that what Tide suggested isn't quite right- you can't "find the time at which he/she lands (y = 0) and use that value of t in the first equation to find v0".

    What you can do is use the length of the jump given and solve the two simultaneous equations
    [tex]v_0cos(23)t= 8.7 [/tex] and
    [tex]v_0sin(23)t- \frac{g}{2}t^2= 0[/tex]
    for both t and v0.
     
  6. Sep 4, 2004 #5
    I'm sorry, I feel retarded but I still don't understand how you can have to unknown variables in one problem. If I don't know Vo or time how do I even solve the first equation?
     
  7. Sep 4, 2004 #6
    you can always remove one variable by substituting it as something else....

    for instance, you can use t = x/(cos23 *v)
     
    Last edited: Sep 4, 2004
  8. Sep 4, 2004 #7

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    Picky, picky! :smile:

    Obviously, when I say "solve for" I meant solve for t in terms of [itex]v_0[itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Kinematics in Two Dimensions?
Loading...