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!

Olympic Jumper

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data
    An Olympic long jumper leaves the ground at an angle of 21.8 ° and travels through the air for a horizontal distance of 8.97 m before landing. What is the takeoff speed of the jumper?

    2. Relevant equations
    Square root of V^2_ox + V^2_oy

    3. The attempt at a solution

    Sqrt of 8.97^2 + (-9.8)2, idk what to do with the 28degree
  2. jcsd
  3. Feb 2, 2009 #2


    User Avatar
    Homework Helper

    Your equation to combine the components of velocity doesn't exactly apply to the data given.

    The problem gives you the distance of the jump. Not the horizontal component of V.
  4. Feb 2, 2009 #3


    User Avatar
    Homework Helper

    This is a projectile motion problem. You have to separate the vertical and horizontal motions because different formulas apply. The horizontal part is motion at constant speed while the vertical part is accelerated.

    Use Vx = v*(21.8), Vy = v*sin(21.8)

    For the horizontal part, use x = Vx*t
    For the vertical part, use V = Vy + at and y = Vy*t + .5at^2
    Put the numbers you know in all three formulas. It should then be possible to solve one of them and find something out (usually the time) so you substitute in another of them and find what you want.
  5. Feb 2, 2009 #4

    Let [tex]O(X,Y)[/tex] be a frame of reference with origin in the jumper. The equations of velocity and space in function of time are:


    v_x=v_o cos\theta


    v_y=v_o sin\theta + g*t


    s_x=v_o cos\theta * t +s_{ox}


    s_y=\frac{1}{2}g*t^2 + v_o * sin \theta + s_{oy}

    I assume [tex]s_{ox}[/tex] and [tex]s_{oy}[/tex] null inasmuch as origin is in the jumper. So the relation [tex]s_y=f(s_x) [/tex] by deleting [tex]t[/tex] is:

    [tex]s_y=\frac{1}{2}g*\frac{s_x^2}{(v_o * sin \theta)^2} + s_x * tan\theta} [/tex]

    By substituting [tex] \theta=21.8°[/tex], [tex]s_x=8.97m [/tex]and [tex]s_y=0 [/tex] you'll get [tex]v_o[/tex]

    I hope this help, bye...:wink:
    Last edited: Feb 3, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Olympic Jumper
  1. Physics Olympics (Replies: 0)

  2. Physics Olympics? (Replies: 1)

  3. Bungee Jumper (Replies: 3)

  4. Ski jumper (Replies: 1)

  5. Jumper Question (Replies: 2)