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Calculating Jump Landing Position

  1. Jul 31, 2009 #1
    Hi,

    I need a bit of help with the following:

    I need to calculate the landing position of a man if he jumped, using his speed, gravity and jump height. This needs to be done in 3D-space.

    I would use the template, but I really don't know where to start. I imagine it's something to do with momentum and such, but I'm really struggling to grasp what I need to do.

    Any sort of help at all would be great!

    Thanks,
    Rich
     
  2. jcsd
  3. Jul 31, 2009 #2

    rock.freak667

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    Why do you need to do it in 3D space for? Is there an associated diagram with the man jumping in (x,y,z) coordinates?
     
  4. Jul 31, 2009 #3
    Assuming drag force from air is minimal to 0, the time it takes for an object to fall is independent of its speed along the x or z axis (w/o air a bullet and a dropped ball will fall with the same speed.)
    Therefore, using his jump height, his initial speed along the y axis and some basic equations, you can figure out how long he'll be in the air. Therefore, by knowing time (t), and the velocities in the x and z direction determine how far he'll go in each direction.

    The problem may be more complex and beyond me (i've only taken AP physics), but i hope this helps.
     
  5. Jul 31, 2009 #4
    Hi,

    Well, that sounds like it's what I want... Can you write the equation I would need? It's easier for me to grasp stuff from equations!

    Cheers,
    Rich
     
  6. Jul 31, 2009 #5
    Well the easiest way to do this is to realize that without air resistance or anything interfering with his jump that the total time of the jump is equal to twice the time it would take him to fall from that height. It takes the same time to go up as it does to come down in this case.

    So we simply calculate the free-fall time from a height 'h' and multiply it by two giving:
    T = 2*sqrt(2h/g)
    where T is the total time he spends in the air (including going up), and g is the positive acceleration due to gravity.

    The rest is easy since we now know the time simply use the equation v = d/t
    and you can see that the position (assuming he lands at a place level with his original surface) is directly in front of him a distance of 'r' :
    r = 2v*sqrt(2h/g)
    v is his velocity before jumping.
     
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