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Speed in Space

  1. Jul 21, 2009 #1
    A projectile is launched vertically from the surface of the Moon with an initial speed of 1210 m/s. At what altitude is the projectile's speed three-fourths its initial value?

    can someone please help me with this!!! i keep getting 1.96x10^5 m but it says that is not the correct answer

    thanks
     
  2. jcsd
  3. Jul 21, 2009 #2
    This should really be in the hwk section. However, yes, your answer is incorrect. Think of it this way:

    [itex]\frac{1}{2}mv_1^2=\frac{1}{2}m (\frac{3v_1}{4})^2+mgh[/itex] (assuming you're allowed to use mgh). I assume you've covered conservation of energy (or are you still only doing kinematics)?
     
  4. Jul 21, 2009 #3
    yes we have covered conservation of energy. however when i solve using this equation i get the answer 195885, which is the same thing as 1.96x10^5 right?
     
  5. Jul 21, 2009 #4
    Well you're making an algebraic mistake somewhere. I'm getting 3.3x10^4
     
  6. Jul 22, 2009 #5

    HallsofIvy

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    Staff Emeritus
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    And, of course, since we are talking about the moon and not the earth, you have to use the value of "g" that is correct for the moon, not 9.81 m/s2.

    What is that "g"?
     
  7. Jul 22, 2009 #6
    1.63 m/s^2
     
  8. Jul 22, 2009 #7
    More precisely, an unconfirmed source says 1.62631, but I haven't done the calculation or found corroboration--but it's close enough to the approximation I do know that I don't doubt it.
     
  9. Jul 22, 2009 #8

    turin

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    You don't need to make an assumption; you can determine this by comparing to the more general expression. However, even if mgh turns out to be a good approximation, you might as well use the more general expression to begin with: U=GMm/r^2.
     
  10. Jul 22, 2009 #9
    once i learnt the conservation of energy...physics became a whole lot easier coz i could basicly solve almost all problems using a few basic energy equations...instead of kinematics...
     
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