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Find Initial Velocity from Angle and Horizontal Distance

  1. Feb 12, 2013 #1
    1. The problem statement, all variables and given/known data
    "An athlete executing a long jump leaves the ground at a 30.0° angle and travels a horizontal distance 8.00m. What was the take-off speed?"

    Let
    ##x=## horizontal distance
    ##v=## velocity
    ##a=## acceleration

    Known:
    ##x=8 m##
    ##x_0=0 m##
    ##a_y=-9.8 m/s^2##
    ##v_{0x}=v_0cos(30°)##
    ##v_{0y}=v_0sin(30°)##

    Solve for ##v_0##


    2. Relevant equations
    The only acceleration present is gravitational acceleration, so constant acceleration formulas are valid here. I decided to work with the horizontal component, because I was given the horizontal distance.
    ##v_x^2=v_{0x}^2+2a_x(x-x_0)##
    ##x=x_0+v_{0x}t+1/2a_xt^2##
    ##x=x_0+1/2(v_{0x}+v_x)t##


    3. The attempt at a solution
    I decided that since the athlete lands, and thus stops moving, at ##x=8,## the ##x## component of the final velocity would be ##v_x=0.## Given this and the values given above, I began plugging my values into the first relevant equation. But that led to ##v_0=0##!

    ##v_x^2=v_{0x}^2+2a_x(x-x_0)##

    ##v_0x=v_0cos(30°).##

    ##0=(v_0cos(30°))^2,##

    ##v_0=0##


    On my next attempt I went out on a bit of a limb and used the third relevant equation to derive time ##t,## which I knew would contain the unsolved variable ##v_0.## I got this and plugged this into the second relevant equation, which made the two ##v_{0x}## variables cancel out.

    ##x=x_0+v_0t+1/2a_xt^2##

    ##t=16/v_0##

    ##x=x_0+v_{0x}t+1/2a_xt^2##

    ##8=0+v_{0x}(16/v_{0x}),##


    There's something I'm not taking into consideration (and I might be making this harder than it needs to be), but I don't know what. Any hints as to where I should be directing my thinking on this?
     
  2. jcsd
  3. Feb 12, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    You are not taking the y displacement under consideration. And there is no x-acceleration. The only accelerative agent is gravity.

    So write the equations for the x and y coordinates.
     
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