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Simple conservation of energy problem

  1. Sep 27, 2008 #1
    1. The problem statement, all variables and given/known data
    All I need to do is figure out at what height h above the ground does a projectile have .5v using only v and g in the answer.


    2. Relevant equations
    Conservation of energy: KEi=GPEf
    .5mv2=mgh


    3. The attempt at a solution
    so I went ahead and just plugged in .5v or v and solved for h so...
    .5(.5v)2=gh
    .5(.25v2)=gh
    .125v2=gh
    h=v2/8g

    however, the correct answer is 3v2/8g

    so I guess all I need to know is where the 3 came from. I don't see any algebraic errors and I frankly don't have any other ideas
    Any help is greatly appreciated,
    Thanks!
     
  2. jcsd
  3. Sep 27, 2008 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi ehilge! Welcome to PF! :smile:
    .5(v)2=gh

    Too clever by half!! :biggrin: :wink:
     
  4. Sep 27, 2008 #3

    Doc Al

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    Staff: Mentor

    You need to set up the conservation of energy equation properly. What's the total energy when the projectile just leaves the ground? What's the total energy when its speed = 0.5v and its height = h? Set those equal.

    In general, this is not true. It's only true if all the KE is transformed into PE.
     
  5. Sep 27, 2008 #4
    ok, so I got the correct answer when I set .5mv2=.5m(.5v)2+mgh. I think I actually just read my own question wrong. I thought I was trying to find the height if the initial velocity is .5v but it looks like I needed to find what the height is when the velocity is .5v but the intitial is still v. Oh well, I geuss I better pay more attention next time. Thanks for you help!
     
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