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2D Kinematics - Dropping object from a zipline

  1. Mar 15, 2016 #1
    1. The problem statement, all variables and given/known data
    A 50m zipline is hung 11m off of the ground at its starting point (assume it ends at ground level). If you are to drop an object which is 20m along the ground from the start of the zipline, how far down the line would you need to let it go?

    2. Relevant equations
    d=((VF+VI)/2)*t
    VF = VI + a*t
    D=(VF*t) + 0.5*a*t2
    Vf2 = VI2 + 2*a*d

    3. The attempt at a solution
    I was able calculate the time as 25s for the entire line, which gives an acceleration of 16m/s/s and a final velocity of 40m/s. Also solved for:
    • Vfx (max) = 39.2m/s
    • Vfy (max) = 8.8m/s
    • ax = 15.7m/s/s
    • ay = 3.52 m/s/s
    At the X=20m position (the target) it would take 15.9s to be directly above it, and that point is 4.4m below the height of the starting point. I also know that the time it takes the projectile to fall from the zip line to the target will also be the time between the release point on the zip line and the point at which you'd be directly over the target.

    What equation do I use to calculate the release point along the zip line? This stuff is easy if the plane of motion is parallel to the ground, but in this case it's a decreasing distance at an angle...
     
  2. jcsd
  3. Mar 15, 2016 #2

    gneill

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    Staff: Mentor

    Your problem statement is not clear. Is the objective to find the point of release along the zipline in order for a falling object to hit a target 20 m along the ground from the start of the zipline? So find distance d in the following figure?
    upload_2016-3-15_15-21-49.png
     
  4. Mar 15, 2016 #3
    Yes, in the diagram above we are solving for distance D along the zip line
     
  5. Mar 15, 2016 #4

    gneill

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    Staff: Mentor

    I don't think that this will be a simple problem if you assume that the object accelerates down the zipline before being released; It will have different initial release speed at every position along the line. At first glance I'd suggest finding the release speed as a function of D via conservation of energy considerations. Then analyze the subsequent projectile motion (you'll have initial speed, direction, and launch height) to find the point of impact with the ground. That is, find X(D), the horizontal position of the impact point. Finding the inverse relationship D(X) may be an algebraic challenge...
     
  6. Mar 15, 2016 #5
    Can you elaborate on "finding release speed as a function of D via conservation of energy considerations"?

    I had somebody else suggest picturing this as a light box model and timing the shadow (i.e. the X-axis velocity) and going from there....which seems similar to what you're suggesting about finding the horizontal position of the impact point (but then again, isn't the horizontal position just 2m?)
     
  7. Mar 15, 2016 #6

    gneill

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    As the object slides along the zipline it loses gravitational potential energy and gains kinetic energy in equal measure. So if you know the drop in height for a given zipline position D then you can determine the kinetic energy of the object at that position. Kinetic energy is proportional to velocity. Hence you can find v as a function of D.
    Your problem statement says (if I've interpreted it correctly) that the target is at 20 m. Note that the x-velocity of your shadow will be accelerating while the object is attached to the zipline, then constant after release. The time to impact after release will depend on the release height and the release velocity y-component, which is also a function of the release position. So you need to find general expressions for the release conditions (x-position, y-position, velocity) and then determine the point of impact. Eventually you should be able to write a single expression for the point of impact X as a function of the release point. Turn that around to find D as a function of X.
     
  8. Mar 15, 2016 #7
    Yes, that should have read 20m, not 2m.... my 0 key requires a little convincing sometimes....

    Regarding the first part there, would v as a function of D be the v(X), v(Y), or v(Combined)?
     
  9. Mar 15, 2016 #8

    gneill

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    Staff: Mentor

    It would be the translational speed along the zipwire, so "v(Combined)". The change in gravitational potential energy becomes kinetic energy and the motion is constrained to follow the zipwire.
     
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