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Speed of a rocket using energy methods

  1. May 22, 2014 #1
    1. The problem statement, all variables and given/known data
    A rocket is launched such that when the fuel is exhausted, the rocket is moving with a speed of Vo at an angle of 37° with the horizontal and at an altitude h. (a) Use energy methods to find the speed of the rocket when its altitude is h/2. (b) Find the x and y components of velocity when the rockets altitude is h/2. Use the fact that Vx=Vxo = constant (since ax-0) and the result from a.


    2. Relevant equations
    The Kinetic and Potential energy equations


    3. The attempt at a solution
    I did
    1/2mv2+mgy= E
    and then I solved for v and got
    v= (2Egy)1/2
    the correct answer is (a) v = (v02 + gh)1/2
    (b) vx= .799vo, vy= - (- 0.362vo2 +gh)1/2
    I wasn't really sure how to approach this type of problem
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 22, 2014 #2

    vela

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    How did you manage to get that? It looks like you didn't do the algebra correctly.

     
  4. May 22, 2014 #3

    BiGyElLoWhAt

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    in a you solved for v wrong. You're attempting to do what they did, you just made a mistake that I think you can find.
    When they say the speed at h/2, do they mean on the way up or down?
     
  5. May 22, 2014 #4
    I'm not sure and what could I do with that information?
     
  6. May 22, 2014 #5

    haruspex

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    You have two instances where you can apply that equation: at the point where the fuel is exhausted, and again when it has descended to height h/2. Which term is the same at both points? What equation does that give you?
     
  7. May 22, 2014 #6
    Is it that they both have the same amount of energy? So would you get something like 1/2mv2+mgy=1/2mv2+mgy?
     
  8. May 22, 2014 #7

    haruspex

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    Yes, plugging in the appropriate values for v and y each side.
     
  9. May 24, 2014 #8
    OK I get it but I am having trouble finding the y component. I did sin 37 to find the y component but it looks like in the answer they gave they have that number squared - (0.362vo2 +gh)1/2. Why is sin theta squared in this equation?
     
  10. May 24, 2014 #9

    SammyS

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    What is ##\ (v_0)_y\ ?\ ## Isn't sin(θ0) a factor of this?
     
  11. May 24, 2014 #10
    The initial speed in the y direction is Vo* sin(37). So shouldn't the equation be .601Vo instead of .362vo?
     
  12. May 24, 2014 #11

    haruspex

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    If a speed is 0.601v, what is the square of that speed?
     
  13. May 24, 2014 #12
    OK I see how they got that answer. Thanks everyone!
     
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