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Physics - position/velocity/acceleration

  1. Jun 7, 2009 #1
    okay, here's the problem.

    A rocket falls from rest a 945m with an initial velocity of zero. Suppose there is still fuel in the rocket, and at what height should the engines be fired again to brake the rocket's fall and allow a perfectly soft landing? Acceleration of the rocket engines is +29.4 m/s^2.

    so basically I want to know when should the rocket engines be fired so the velocity can be decreased to zero by the time the it hits the ground.

    I don't know where to start.
     
  2. jcsd
  3. Jun 7, 2009 #2

    PhanthomJay

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    For starters, try using the kinematic equations that relate velocity with acceleration and distance. Write them for each part of the motion. You should get 2 equations with 2 unknowns from which you can solve for the height. Please show your work so we can assist further.
     
  4. Jun 7, 2009 #3
    what?? there are two parts to this motion?
     
  5. Jun 7, 2009 #4

    PhanthomJay

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    Yes, the first part being while the rocket is in free fall until the engines are fired up, and the second, after they are fired. There are different accelerations during each part.
     
  6. Jun 7, 2009 #5

    tiny-tim

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    Welcome to PF!

    Hi johnnyies! Welcome to PF! :smile:
    Yes!!

    First with acceleration g downward, then with acceleration (29.4 - g) upward :wink:
     
  7. Jun 7, 2009 #6
    well, so the acceleration in free fall is -9.80 m/s^2 and the rockets firing up at 29.4 m/s^2.

    How do I figure out the velocity at that point? cause by that point it has to reach some # velocity which must reach zero in the second motion.
     
  8. Jun 7, 2009 #7

    PhanthomJay

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    Repeat, in case you missed it.
     
  9. Jun 7, 2009 #8
    explain further, since it's not helping.
     
  10. Jun 7, 2009 #9

    tiny-tim

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    Call the height at which the engines fire, h …

    then calculate the speed at h, and at the ground, using the standard constant acceleration equations :wink:
     
  11. Jun 7, 2009 #10
    While it would probably be best to use the kinematic equations just to become more comfortable with them, the velocity achieved is not as necessary to know as the net accelerations experienced by the rocket. It becomes more or less a critical thinking / logic problem.

    Some points for consideration:
    (1) If the acceleration gravity is -10 m/s^2 and the acceleration of the rocket's thrust is +30 m/s^2, the competing net accelerations are -10m/s^2 and +20 m/s^2 respectively.

    (2) The distance and velocity plotted will appear as a parabolic curve, velocity begins at 0 m/s (at 945 m), peaks at some unknown height 'h', and ends at 0 m/s (at 0 m).

    (3) At the unknown point, thrust is initiated and the +20 m/s^2 acceleration gradually overcomes the accumulated velocity from gravity.

    (4) Logically, it will take twice the distance for the rocket to achieve peak velocity as it will take to reduce the velocity to zero comfortably.

    Sometimes restating the known in a more simple way can shed some insight into the solution. That's why drawing a sketch and writing all known values for common variables is the first step in approaching a physics problem. The forces in this problem can be simplified to the values I used because 29.4 = 9.8 * 3, which is of no coincidence by the author. This is a problem that is very easily made complicated but with some familiarity the problem can be solved quite rapidly. As such, it is a staple for exams like the MCAT.

    It's best to struggle through these kinds of problems and carefully disassemble them. The process of understanding imparts upon you a new perspective and approach that you otherwise might not have considered.
     
    Last edited: Jun 7, 2009
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