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Bullet hits the ground on its return

  1. Feb 7, 2012 #1
    V^2=Ae^-2kx - g/k (upward motion)
    V^2=g/k - Be^2kx (downward motion)

    Use the above result to show that , when the bullet hits the ground on its return, the speed will be equal to the expression

    Vo Vt / (Vo^2 + Vt^2)^0.5

    In which Vo is the initial upward speed and Vt= (mg/C2)^0.5= terminal speed = (g/k)^0.5

    This is result allows one to find the fraction of the initial kinetic energy lost through air friction

    THIS PROBLEM WILL KILL ME I SPEND DAYS TO SOLVE IT BUT I COULD NOT:(
     
  2. jcsd
  3. Feb 7, 2012 #2
    Can you put this question in a more formatted manner? It's rather hard to follow which is probably why people are viewing but not repsonding...
     
  4. Feb 8, 2012 #3
    this is may be more clear.
    thanks.
     

    Attached Files:

  5. Feb 8, 2012 #4
    did you solve question 2.12? if so what was your answer? 2.12 didn't come out clear enough for me to solve on my own. The question says to use that answer (not the equations) to solve it...
     
  6. Feb 8, 2012 #5
    Yes i did 2.12 .. I proved *the equations*!
     
  7. Feb 8, 2012 #6
    Please can any body help me!!!
     
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