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A rocket-powered hockey puck

  1. Mar 12, 2009 #1
    1. The problem statement, all variables and given/known data
    A rocket-powered hockey puck has a thrust of 2.40 N and a total mass of 2.40 kg. It is released from rest on a frictionless table, 3.80 m from the edge of a 2.30 m drop. The front of the rocket is pointed directly toward the edge.

    How far does the puck land from the base of the table?



    3. The attempt at a solution
    in solving the problem, this is what i got:

    2.40 N = 0+1/2(9.8)t^2
    4.80=9.8*t^2
    sqrt(0.49)=t
    t=0.70 s

    F=ma
    240N=240kg*a
    a=1 m/s^2

    v_x^2-v_o^2=2aS
    v_x^2=2(1 m/s^2)(3.80 m)
    sqrt(2*1*3.80)
    v_x=2.76 m/sec

    x=(2.76 m/sec)(0.70 sec)
    x= 1.93 m

    I plugged that in as the answer but i was told that it is wrong. Can someone please check my work? i'd really appreciate it :D

    thanks in advance :D
     
  2. jcsd
  3. Mar 12, 2009 #2

    lanedance

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    Homework Helper

    what is this this equation? Force does not equal 1/2. a.t^2 looks like a distance

    do you mean 2.3m for the puck to fall from the table?
    horizontal acceleration of the puck - looks good
    velocity at edge of table due to rocket...
    does the puck keep accelerating horizontally after it leaves the table?
     
  4. Mar 12, 2009 #3
    yeah i made a mistake with the number. i got 0.685 sec for the time when i plugged 230 m in instead of 2.40 N.

    for that answer, i got 1.90 meters but i was still wrong.
    as for accelerating horizontally, wouldn't it be accelerating vertically instead of horizontally?
     
  5. Mar 12, 2009 #4
    mistake: i didn't use the whole equation to figure out the horizontal acceleration.
    i should have used this equation:
    x=v_f*T +1/2 at^2

    thanks for all the help. i really appreciate it :D
     
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