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Initial velocity required for object to have x ke at y height

  1. Aug 8, 2013 #1
    ... factoring in fluid dynamics, not just Newtonian physics...

    At a given height, the object will possess 75% of the kinetic energy of when it initially began moving.
    Mass is irrelevant.

    Given variables:

    -Drag coefficient
    -Fluid density
    -Cross-sectioned area
    -Displacement
    -Initial and final kinetic energy (sort of; there's not exact value, but I'm just using 1 for initial and 0.75 for final)
    -Acceleration/deceleration due to gravity
    -etc

    Missing variables:

    Time and initial velocity.

    Relevant equations would be the kinetic energy equation, drag equation, and deceleration from force, as well as some basic kinematics ones.

    I'm trying to create a differential equation where time is isolated on one side while velocity is removed, but I'm having trouble.

    If anyone could post the differential equation, with proper steps for derivation, I'd appreciate it :smile:
     
    Last edited: Aug 9, 2013
  2. jcsd
  3. Aug 8, 2013 #2

    SteamKing

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    Would you like fries with that?

    I'm sorry, PF rules require you to show some effort in order to receive help. We don't take requests to provide solutions.
     
  4. Aug 9, 2013 #3
    Yes, fries would be good.

    Anyways, I've tried deriving it myself, but *cough* I'm embarrassed to admit it, but I got stuck deriving the kinematics part >.>

    [itex]2Vi*t+\sqrt{3}*Vi*t-\sqrt{3}*a*t^2-d=0[/itex]
     
    Last edited: Aug 9, 2013
  5. Aug 14, 2013 #4
    Tried to calculate what's required to reach that height first, but I don't know how to work with when a variable depends on itself.

    [itex]V=V_i-((\rho*C_d*A*V^2)/2M - G)T[/itex]

    [itex]D=\int_0 (V_i-((\rho*C_d*A*V^2)/2M - G)T)dT[/itex]
     
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