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PLEASE:Projectile motion quicky problem with web assign

  1. Oct 11, 2007 #1
    Okay. So I have this problem that I posted last night. https://www.physicsforums.com/showpost.php?p=1462108&postcount=1 I have solved everything.

    My problem is that I entered the value of Maximum height above the cliff that I got which was 439.5meters into WebAssign (online homework). The problem is that I am quite positive that the answer should be 439.5-205, because it asks for the height ABOVE the cliff top.

    But WebAssign accepted the answer that I used (439.5)

    Am I correct that it should be the difference since I used [tex]v_y^2=(v_0)_y^2+2a\Delta y[/tex]

    so[tex]0=(155sin37)^219.6\Delta y[/tex] so[tex]\Delta y=439.5[/tex]

    I am tutoring tonight so I really would like to give this person an unambiguous answer!
     
  2. jcsd
  3. Oct 11, 2007 #2
    If someone could just verify that my concept is correct, that is all I need.

    I am convinced that Max height above the cliff is (Delta y)-(cliff height).
     
  4. Oct 11, 2007 #3

    learningphysics

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    The height above the top of the cliff is 439.5m

    The height above the ground is 439.5m+205m.
     
  5. Oct 11, 2007 #4

    learningphysics

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    vf^2 = vi^2 + 2ad

    In this formula d is the displacement between the position where the object is at vi, and the position where the object is at vf... at vi is the top of the cliff... at vf it is at the maximum height. so d here is the height from the top of the cliff to the max...
     
  6. Oct 11, 2007 #5
    Thanks you! I am pretty sure I am with you now...

    By plugging in v_f=0 you are "telling" the equation that displacement is from where the projectile was fired to the point where v_f=0. I was just confused as to how it does not "assume" that I mean when it lands...v_f=0 when it lands too doesn't it? I know I am confusing a concept here....I have done this before with this same kind of problem a couple of semesters ago...

    Casey
     
    Last edited: Oct 11, 2007
  7. Oct 11, 2007 #6

    learningphysics

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    exactly.
     
  8. Oct 11, 2007 #7

    learningphysics

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    No it isn't 0 when it lands.
     
  9. Oct 11, 2007 #8
    So when it lands, it still has a velocity at that instant.

    When it comes to rest completely (immediately after landing) we are no longer dealing with projectile motion, correct?

    We would use v_f from the projectile motion problem to make any further presumtions with regards to after its landing i.e, kinetic energy, impact-momentum, Force, etc...
     
  10. Oct 11, 2007 #9

    learningphysics

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    exactly.

    yes... it becomes a different problem once it hits the ground... the projectile motion is from the moment it is released to right before it hits the ground...
     
  11. Oct 11, 2007 #10
    Understood. Out of curiosity now; Suppose I set up an equation that I wished to solve for v_f not equal to zero...at what time after being launced will a projectile's vertical component of velocity be V. Now this velocity v_y=V will occur twice correct? On the way up and on the way down. Now the equation [tex](v_y)_f=v_o+a_yt[/tex], will that generate the upward or the downward? Or will depend on a sign?

    This may be a stupid question:rolleyes:..I have to think about it:smile:
     
  12. Oct 11, 2007 #11

    learningphysics

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    yes, it's 2 different values... suppose V is positive... so going upward vy = V... downward vy = -V. so you'll get two different times...

    Now, the height at which these 2 velocities occur is the same. (think conservation of energy)...

    This can be seen by the equation vf^2 = vi^2 + 2ad

    it doesn't matter if vf is +V or -V... you'll get the same answer for d.
     
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