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Physics highest point reached by skateboarder

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data

    A skateboarder shoots off a ramp with a velocity of 6.5 m/s, directed at an angle of 55° above the horizontal. The end of the ramp is 1.5 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp.

    (a) How high above the ground is the highest point that the skateboarder reaches?

    (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

    2. Relevant equations



    3. The attempt at a solution

    (a) i figured i need y+1.5 for my answer so i need to find y.
    so sin55= y/6.5
    y= 5.32
    y+1.5 = 6.82 but its wrong

    (b) i thought it was as easy as using the phythagoram therom and said a^2 + 5.32^2 = 6.5^2
    a= 3.73 but thats wrong

    i inserted the pictures i drew for each part.
     

    Attached Files:

  2. jcsd
  3. Sep 17, 2009 #2

    kuruman

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    First write down the kinematics equations for two-dimensional projectile motion. If you are not sure what to do with them, we will help along, but we need something to point to.
     
  4. Sep 19, 2009 #3
    ok y=y_0 + v_0*t-g/2*t^2

    now what do i do?
     
  5. Sep 19, 2009 #4
    That's not the kinematics equation that we want/require, since it involves time, which does not feature in part(a) of the question.
    For part (a), what is the final velocity (y-component) of the skateboarder at the highest point?
    For part (b), what is the time taken for the skateboarder to reach this highest point? What is his velocity in the x-direction throughout the trajectory?
     
  6. Sep 19, 2009 #5
    ok the only one that doesnt involve time is y-y_0=V_0y/V_0x * (X-X_0) - g(x-X_0)^2/ 2*V^2_0
     
  7. Sep 19, 2009 #6

    kuruman

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    There is another one that does not involve time and has no "x" subscripts.
     
  8. Sep 20, 2009 #7
    ok well im looking at my formula sheet given to me by the instructor and the only other formula on there that does not have t or an x subscript is V^2- V_0^2 = 2a *(x-x_0)
    and that formula is not even on the chapter we are in.
     
  9. Sep 20, 2009 #8

    kuruman

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    Use this for the motion in the y-direction, but first you need to figure out what to put in for these quantities - which ones you know and which one you are looking for. Any ideas?
     
  10. Sep 20, 2009 #9
    ok well the initial velocity would be 0. v= 6.5, X=1.5 so we dont know a or X_0
     
  11. Sep 21, 2009 #10

    kuruman

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    You know a - don't forget that once he gets airborne he is in free fall. You may not know x_0, but you don't care what it is. You are looking for his vertical displacement which is x - x_0. So ...
     
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