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Homework Help: Variable Problem for 2D Kinematics

  1. Jan 27, 2008 #1
    Here's another variable problem

    1. The problem statement, all variables and given/known data
    An extreme sportsman jumps from one cliff to another of width (w) by leaping horizontally by vo. With a height difference of h (where the starting cliff is higher) what is vo so he doesn't miss the cliff? (also, what is the direction of motion)

    2. Relevant equations
    x=w, y=0


    3. The attempt at a solution
    Determine time of landing:

    Determine vo:

    does this look correct so far?

    I'm stuck at finding the direction of motion, i know the equation is:

    =tan[tex]^{-1}[/tex] (2h/w) degrees of the horizontal component

    but, i'm not sure if this is exactly right...
  2. jcsd
  3. Jan 27, 2008 #2

    Shooting Star

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    How much time does he take to cross a horz dist of w, if he starts off with a horz velo of v0? Equate that time with the time you have got.

    The final dircn is given by tan(theta) = vy/vx, at that point.
  4. Jan 27, 2008 #3
    Thank you for your response.
    I was wondering if you could please elaborate on your post.
    My work is in my original post where i substituted time into the vot equation. Is this what you meant by equating time with the time that i have? (not quite sure what you mean)

    also, i have used that tan ratio for direction of motion. does my resultl look reasonable? or is there a way i can simplify the variables?
  5. Jan 27, 2008 #4
    The time to make the jump: Solve for t in: h = ½*g*t^2.
    And then solve for v0 in: s = v0*t.

    For the direction, just make the two values an vector, and determine the angle.
  6. Jan 28, 2008 #5

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    It is absolutely correct. Somehow, I overloked this part. Sorry for that.

    Where's the minus sign of (-gt) gone? The value is OK. The direction is given by tan(theta) = -2h/w, where theta is the angle the velo makes with the x-axis, that is, the direction of v0. Since the body is moving downward and to the right, the angle is negative.
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