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Projectile's initial speed?

  1. Jun 23, 2009 #1
    A projectile launched vertically from the surface of the Moon rises to an altitude of 270 km.
    What was the projectile's initial speed in m/s?

    okay so I keep getting this answer wrong but Im pretty sure Im doing it correctly.....

    this is what I am doing:

    2*6.67*7.35*270/3.02 and then all of that multiplied by 10^2 and then that answer square rooted and I get 936.27 m/s .....but that is not correct...HELP! what am I missing??
     
  2. jcsd
  3. Jun 23, 2009 #2

    CompuChip

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    Can you maybe explain where you got all those numbers from?
    What is 6.67, what is 7.35, what is 3.02, where do the 2 and 10^2 come from and why square-root it?

    i.e. post your formula and identify the variables. Now it's just a string of numbers without physical meaning.
     
  4. Jun 23, 2009 #3
    Vi- square root 2GMmH/(1.74*10^6)2

    7.35 *10^22 kg is the mass of the moon Mm
    Radius of the moon Re=1.74*10^6
    g= 6/67*10^-11
    height reached by the projectile h= 260km = 270* 10^3
    I plugged all of that in to the equation above and got the answer previously stated but it is wrong.
     
  5. Jun 23, 2009 #4
    sorry Vintial = **
     
  6. Jun 23, 2009 #5
    Sounds like a kinematic equation problem to me.

    qxnyfl.gif
     

    Attached Files:

  7. Jun 23, 2009 #6

    rl.bhat

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    Apply the conservation of energy theorem. When the projectile reaches the maximum height, the KE of projectile is equal to the change in PE.
    PE on the surface of the earth = GMm/R. When it reaches the maximum height, the PE = GMm/(R+h)
     
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