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Tricky one

  1. Oct 29, 2007 #1
    A fireworks rocket is launched vertically upward at 40 m/s. At the peak of its trajectory, it explodes into two equal-mass fragments. One reaches the ground t1= 2.51s after the explosion.
    When does the second reach the ground?
    t= ......??

    logiclly, it is 2.51 s also ....but I don't know. so could you pls guys help with that out ??
     
  2. jcsd
  3. Oct 29, 2007 #2
    Ok, find where it is at its peak. At its peak its velocity is zero. Use conservation of momentum to get the initial directions and speeds. That is, let the first fragment travel in some initial direction with an angle theta to the horizontal (or whatever you like), then get the other fragment initial direction and speed, and solve the equations of motion.
     
  4. Oct 29, 2007 #3
    thnx dude for your post. but could you pls explain a little bit more for me?
     
  5. Oct 29, 2007 #4
    Ok, use the equations of motion to find where (at what height) the rocket reaches its peak. Then define a set of axes: to the right horizontally is you +x axis, and, say, vertically upwards is your +y axis. Then say immediately after the explosion, the first fragment leaves at an angle theta relative to the +x axis (in the usual way we measure angles, in the anti-clockwise direction), with a speed v.
    Then you can use the equations of motion to find this angle theta and the speed v, since we know it took 2.51 seconds to reach the ground.
    Now we have the initial direction and speed the first fragment had. Also, initially, linear momentum was conserved, so the other fragment flew off in the opposite direction but with the same speed (because their masses are the same). You know the angle the first one flew off in, so you can find the angle the other one flew off in, because the total momentum initially must have been zero.
     
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