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Slope and projectile problem

  1. Jan 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Eddie "The Eagle" Edwards takes off on the 90 m ski-jump with a horizontal velocity of 20 m/s. If the gradient of the landing slope is as shown, find the distance L along the slope covered by his jump,ignore any effect of the air on this eagle.

    Image given:
    http://s1176.beta.photobucket.com/user/LolaGoesLala/media/yeahok_zps7743bc20.png.html

    3. The attempt at a solution
    So, they have given us the horizontal velocity which is 20 m/s [ E ]
    and i am not sure what the 90 m represent.

    Is it the verticle length or is the diameter along the base of the triangle.
     
  2. jcsd
  3. Jan 4, 2013 #2

    jedishrfu

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    He started from zero velocity travelledsome dist then jumped with a horizontal vel of 20m/sec and the diagram shows the angle of the skijump.

    Since he fell some distance and gained kinetic energy you should have enough info to solve the problem.
     
  4. Jan 4, 2013 #3

    gneill

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    I think that you can ignore the "90 m"; consider it a name of the jump.

    Concentrate on the fact that Eddie is launched horizontally, and that you are given information about the gradient (the "slope" of the slope, so to speak).
     
  5. Jan 4, 2013 #4

    haruspex

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    From a little Googling, the 'size' of a ski jump is to do with distance from take-off to some standard landing line, rather like par for a golf hole. If so, it doesn't supply any useful info here. So it would seem the question ought to say "with a velocity of 20 m/s horizontal", i.e., there is no vertical velocity initially. (In practice, the jump slope is a little below horizontal.)
     
  6. Jan 4, 2013 #5
    Okay.
    so i know the horizontal velocity = 20 m/s
    θ = E 37 S

    so i think i would be able to use the conservation of energy in this question.
    mgh = 1/2mv^2
    gh = 1/2v^2
    (9.8m/s^2)h = 1/2(20m/s)^2
    h = 20.4 m

    so can i use sin 37 = 20.4 m / L
     
  7. Jan 4, 2013 #6

    gneill

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    The 20 m/s is a horizontal velocity. The "h" in "mgh" pertains to a vertical displacement. So they are unrelated. Conservation of energy isn't going to cut it here.

    Instead, look at Eddie's trajectory as a function of horizontal displacement and the line of the ski slope as a function of horizontal displacement.
     
  8. Jan 4, 2013 #7
    well i was looking over my notes.
    and you are right.
    But i also found this equation
    distance horizontal = v^2 sin 2θ / g
    but doesn't that only work for like the velocity that includes both verticle and horizontal
     
  9. Jan 4, 2013 #8

    gneill

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    That would be the range equation for a projectile launched with speed v at angle θ over a horizontal surface. That, too, doesn't pertain to the given situation. I doubt that you'll be able to pull a "canned" formula out of your pocket to fit this situation. Instead, you'll have to develop a method of solution.

    If you step back from the problem and look at your diagram, you should be able to see that there are two functions and a point of their intersection that you're interested in. One is the parabolic trajectory of the skier, the other is the straight line of the ski slope.

    If you set your origin at the launch point, can you write an equation expressing the line of the hill as a function of horizontal distance? How about the vertical position of the skier as a function of horizontal distance?
     
  10. Jan 4, 2013 #9
    I am sorry but i am confused by your two question.
     
  11. Jan 4, 2013 #10

    gneill

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    attachment.php?attachmentid=54465&stc=1&d=1357358573.gif

    Two functions that have a point of intersection.
     

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  12. Jan 4, 2013 #11
    So your first question is about the slope (blue line) and the x distance ?
     
  13. Jan 4, 2013 #12

    gneill

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    Yes. Write an equation for it given the information that you have.
     
  14. Jan 4, 2013 #13
    So like...
    well the distance is just east while the blue is east 37 south to the distance horizontally
     
  15. Jan 4, 2013 #14

    gneill

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    Hmm. Have you taken a math course in functions? Equations of lines, parabolas, and so on?
     
  16. Jan 4, 2013 #15
    yeah. but isn't this physics thought? :O
     
  17. Jan 4, 2013 #16

    gneill

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    The language of physics is mathematics. It provides the tools you need to solve problems.

    Here you have an opportunity to use what you learned in your functions course to solve a physics problem. All the math you've learned will come in handy as you go forward. The trick is to recognize its applicability to a given problem :wink:
     
  18. Jan 4, 2013 #17
    Oh i see..
    But i don;t get it im i supposed to use the y =mb+x functions?
     
  19. Jan 4, 2013 #18

    gneill

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    Yup. The ski slope can be represented by a straight line. The trajectory of the skier is a parabola. They intersect at a point. Physics is essentially the world described in terms of mathematics.
     
  20. Jan 5, 2013 #19
    okay so if they have a point of intersection that would mean they must be equal to one another so.

    if the two equations are:
    y = mx

    and the other one is y = -ax

    combing them would be

    mx = -ax

    right?
     
  21. Jan 5, 2013 #20

    gneill

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    That is correct
    Not quite. One is a straight line (y = mx), but the other one is a parabola.

    For the first you have to find the slope m. For the second, you need to find the equation of the parabola that matches the trajectory of Eddie.

    Hint: The equations of motion for a projectile are those of a parabola in what is called "parametric form", where time "t" is the parameter. You will have to convert this to Cartesian form (y as a function of x).
     
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