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Homework Help: Determining Distance

  1. Dec 30, 2007 #1
    1. A 7.1 kg firework is launched straight up and at its maximum height 29 m it explodes into three parts. Part A(1 kg) moves straight down and lands 0.28 seconds after the explosion. Part B(2.5 kg) moves horizontally to the right and lands 12.5 meters from Part A. Part C moves to the left at some angle. How far from Part A does Part C land(no direction needed)?

    2. x-x_0=v_0*\cos{theta}*t, y-y_0=v_0*\sin{theta}*t+0.5*a*t^{2}
    momentum is conserved

    3. I first figured out the initial velocity of A, by using the second equation above, namely -29=v_0*sin(-90)*0.28+0.5*0.28^2, and solved for v_0 of a
    Then I solved v0 of B, from the first equation above, namely 12.5=v_0*cos(0)*0.28
    Then I could solve for the initial velocity of part C and the direction, by using vectors on the momentum.
    Then I can figure out the distance from C to A by the two formulas above, but I keep getting the wrong answer.

  2. jcsd
  3. Dec 30, 2007 #2


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    Do you mean this "-29=v_0*sin(-90)*0.28+0.5*(-9.8)*0.28^2"? If you did that then you should have gotten the right v0. As for B, you can't assume it's time of fall is 0.28sec. It only has a horizontal initial velocity, no vertical component. You can find it's time of fall with the same approach as A. Now use that time of fall to find it's initial horizontal velocity. Now you know everything about the initial velocities of A and B. Use conservation of momentum to find the initial velocity of C.
  4. Dec 31, 2007 #3
    Okay, so here is what I have

    For A, -29=-v0*.28-4.9*.28^2, which implies that v0=102.19994286 m/s

    For B, -29=-4.9t^2, because it is going horizontally, so t=2.432769481 s.
    Thus, 12.5=v0*t, because cos0=1, so v0=5.138176921 m/s, for B

    Then this means that the momentum vector for A is (0,-102.1994286), and for B is 2.5(5.138176921,0)=(12.8454423,0)

    Hence the momentum vector for C is (-12.8454423,102.1994286), and so the velocity vector is 1/3.6*(-12.8454423,102.1994286)=(-3.568178417,28.38873017).

    Then, we know for C, that y-y0=v0*sin(theta)*t-4.9t^2, so -29=28.38873017t-4.9t^2, so t=3.4222779 seconds.

    Then x-x0=v0*cos(theta)*3.4222779=-3.568178417*3.4222779=-12.21129814.

    However, this is wrong. Could someone help in where this is wrong?
  5. Dec 31, 2007 #4


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    Well done! The only problem I can see is when you said you solved "-29=-v0*.28-4.9*.28^2", it looks like you actually solved "-29=-v0*.28-4.9*.28^2". I'm getting about 104.9m/sec for v0.
  6. Dec 31, 2007 #5
    Sorry, but I can't see the difference in the equations. you mean -29=-v0*.28+4.9*.28^2, right.

    But i don't understand why it would be +4.9 and not -4.9.

  7. Dec 31, 2007 #6


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    Well, doing it all again it looks like that was my mistake, not yours. But how about this one. "-29=28.38873017t-4.9t^2". Do you still get 3.422sec? I don't. Unless I'm making another dumb mistake.
  8. Dec 31, 2007 #7
    When I do it, I now don't get it, i don't know why.

    I get that t=6.6796482 seconds, so that should make it work.

    Indeed, that is the correct answer, thanks, problem solved.
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