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Rocket Height-Kinematic Question

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A rocket, initially at rest on the ground, accelerates straight upward from rest with constant net acceleration a, until time t1, when the fuel is exhausted.

    Find the maximum height H that the rocket reaches (neglecting air resistance).

    Express the maximum height in terms of a, t1, and/or g. Note that in this problem, g is a positive number equal to the magnitude of the acceleration due to gravity.

    3. The attempt at a solution

    Okay, in this problem, i am pretty much stuck. I only did the hints that Mastering Physics suggested I do to help me solve this problem.

    I found the height, H1, above the ground at which the rocket exhausted its fuel

    H1=1/2at1^2

    I then found the y velocity, Vy1, that the rocket has when the engine runs out of fuel

    Vy1=at1

    Where do I go from here?
     
  2. jcsd
  3. Sep 14, 2009 #2
    So you got the initial velocity when the rocket runs out of fuel (stops accelerating up) so it is now in free fall. What is the velocity of the rocket when it reaches its max height? Then use the kinematic equation that does not involve time but has displacement in the equation. Essentially the last part of the trip is just like throwing a rock up in the air. You got the velocity upon release, and at max height the velocity is zero.
     
  4. Sep 14, 2009 #3

    Doc Al

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    Staff: Mentor

    Once the rocket runs out of fuel it can be treated like any other projectile in freefall. You have its speed when it runs out of fuel, so figure out how much further it rises.
     
  5. Sep 14, 2009 #4
    Okay, the equation which pgardn suggested is

    Vf^2 = Vi^2 + 2ad


    The velocity of the rocket at the max height of it's path is 0m/s

    Acceleration during free fall equals -9.81m/s^2

    I am assuming the Final Velocity (Vf) is 0m/s, so I set it up like this

    (0m/s)^2 = Vi^2 + 2ad

    Previously, I found the Y velocity which was Vy1 = at1. I use this as a substitute


    (0m/s)^2 = (at1)^2 + 2(-9.81m/s^2)d

    0-(at1)^2=-19.62m/s^2*d

    -(at1)^2
    ------------- = D ?
    -19.62m/s^2


    In terms of a, t1, and/or g, it would be

    H = -(at)^2
    -------- ?
    - 2g
     
  6. Sep 15, 2009 #5

    Doc Al

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    Staff: Mentor

    That's the additional height the rocket will reach measured from the point where it ran out of fuel. What's the total height above the ground? (And please cancel those minus signs.)
     
  7. Sep 15, 2009 #6
    So, previously, I found the height, H1, above the ground at which the rocket exhausted its fuel

    H1=1/2at1^2

    and also using my last equation

    H = (at)^2
    --------
    2g



    The total distance covered by the rocket would equal the sum of these two equations

    H=1/2at1^2 + (at1^2/2g) ?
     
  8. Sep 15, 2009 #7

    Doc Al

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    Staff: Mentor

    Looks good to me.
     
  9. Sep 15, 2009 #8
    Mastering Physics told me this answer was incorrect because

     
  10. Sep 15, 2009 #9

    rl.bhat

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    Homework Helper

    It should be
    H = 1/2*at12 +1/2 (at2)2/g
     
  11. Sep 15, 2009 #10

    Doc Al

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    Staff: Mentor

    You misplaced your parentheses. It should be:
    H=1/2at1^2 + (at1)^2/2g

    (Sorry for not catching that earlier.)

    Use subscripts/superscripts for further clarity:
    H=1/2at12 + (at1)2/2g
     
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