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Rocket Propulsion Speed of Fragments

  1. Oct 24, 2007 #1
    A fireworks rocket is fired vertically upward. At its maximum height of 75.0m , it explodes and breaks into two pieces, one with mass = 1.25kg and the other with mass = 0.240kg . In the explosion, 910 J of chemical energy is converted to kinetic energy of the two fragments.

    a) What is the speed of each fragment just after the explosion?

    b) It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored.
  2. jcsd
  3. Oct 24, 2007 #2
    i have no clue.....omg...my head hurts trying to understand this problem..ahhaahah
  4. Oct 24, 2007 #3
    Wow... Idiot. It's going at a certain speed and the distance between them is a certain number. Which I have no idea of.
  5. Oct 24, 2007 #4


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    Staff: Mentor

    Does the rocket have zero velocity at maximum height?

    a. There is some chemcial energy, 910 J, that becomes kinetic energy of the two fragments. Also the energy is divided according to the momentum, so used the conservation of momentum.

    b. For each piece to hit the ground at the same time, they must have the same initial vertical speed, which has to be _________, based on the momentum at maximum height.

    So each piece is in free fall.


    More generally - http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html
  6. Oct 24, 2007 #5
    so the 910 J is the KE

    and if so...........KE = (0.5*1.25*(v1^2)) + (0.5*.240*(v2^2)) ?
  7. Oct 24, 2007 #6
    i dont kno if the rocket has zero velocity at maximum height...it didnt say in da problem...
  8. Oct 24, 2007 #7


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    Staff: Mentor

    Well if it was traveling upward (vertically) then by definition maximum height is achieved when the vertical velocity is zero (change in vertical displacement has to be zero). So assum the rocket velocity is zero when it explodes.
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