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Using kinematics to find motions with constant acceleration

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data

    painter is standing on scaffolding that is raised at constant speed. As he travels upward, he accidentally nudges a paint can off the scaffolding and it falls a distance 16.0 m to the ground. You are watching, and measure with your stopwatch that it takes a time of 3.26 s for the can to reach the ground. Ignore air resistance.

    What is the speed of the can just before it hits the ground?

    Another painter is standing on a ledge, with his hands a distance 4.35 m above the can when it falls off. He has lightning-fast reflexes and if the can passes in front of him, he can catch it. Does he get the chance?

    2. Relevant equations

    v=vo+at

    y-y0=v0t+.5at2


    3. The attempt at a solution

    I received 11.1 m/s using the second equation and then plugged that number into the first one for an answer of -20.8 m/s.

    Is this right? Thanks!

    I am still working on the second question.
     
  2. jcsd
  3. Sep 13, 2011 #2
    your initial velocity will be the constant speed being lifted up which you don't know yet.

    Its constantly being affected by gravity.

    so, Vf=V0 - gt

    and Vf^2=V0^2 + 2gy

    but velocity just before it hit the ground at the instant is dy/dt=16meters/3.26seconds

    It doesn't seem as though you'd need to find V0, but you might for teh second part of the question. Let me check though. Hold on. God damn. I don't understand the second part of the question. if he is on a ledge but also above the can, how can he also catch it if the can will always be below? o_O
     
  4. Sep 13, 2011 #3
    Oh, I did the first part wrong then?

    I have a solution for the second problem, but if my first part is wrong, then it is wrong.

    Ya, I don't really get the 2nd question either.
     
  5. Sep 13, 2011 #4
    If I were you, I'd just say it is impossible, because he'd have to be below which is impossible.
     
  6. Sep 13, 2011 #5
    As much as I would like to agree(based on common sense?), I received an answer that he can, but only based on if -20.8 m/s is correct.
     
    Last edited: Sep 13, 2011
  7. Sep 13, 2011 #6
    show me exactly how you got 11.1m/s.
    according to my calculation, 11.1m/s is actually the initial velocity; the constant velocity the scaffolding rose.
     
  8. Sep 13, 2011 #7
    Ya, I then use the initial velocity to find for v(t) which using the first equation I obtained -20.8 m.

    Then, for b, the answer should be above 4.35 meters I think, so I just use v2=vo2+2a(yf-yo)
    Which equals 6.28 meters.
     
  9. Sep 13, 2011 #8
    oh ****! excuse my language i think youre right, but...sorry let me think for a few minutes on this.
     
  10. Sep 13, 2011 #9
    haha no problem. I definitely appreciate the help!
     
  11. Sep 13, 2011 #10
    I think the velocity at the instant is actually vy(3.26)=1-1.1 + -19.81(3.26)
    Im really sorry about the first answer I gave you that was completely incorrect. I thought it was right. I have to figure out a way to prove how incorrect it is.
     
  12. Sep 13, 2011 #11
    forgive me its not 1-1.1 its -11.1m/s
     
  13. Sep 13, 2011 #12
    nevermind
     
  14. Sep 13, 2011 #13
    oh wait then 11.1 is actually positive. sorry man im screwing up today. Let me show you more. hold on.
     
  15. Sep 13, 2011 #14
    ok haha. =)
     
  16. Sep 13, 2011 #15
    ok so I have a big brain fart right now. You'll really have to excuse me. This comes to me pretty fast but this question is worded strangely. they said it falls a distance of 16 meters, but it also rises from the point that it will start falling, which suggests that the 16/3.26m/s might actually be right.

    if you want to know how much time it took to fall 16 meters, you'll probably have to assume that since it rises then goes to a velocity of 0m/s, you can consider that your initial velocity from the time its 0m/s to the time it goes down. So, for that times it takes

    16=1/2gt^2

    t=1.8 seconds. which means that the paint thing rose for 3.26-1.8 seconds.

    knowing this is pretty important because then 0=v0-gt when velocity is zero

    so, V0=14.3m/s
    god I might be overanalyzing this, but it seems right now.

    But I still am very sceptical about the velocity
    im pretty sure now that its just gah idk i hope I helped though. what do you think?
     
  17. Sep 13, 2011 #16
    I found this site with a problem very similar to this one. I don't think the can rose but I don't really know. In this similar situation, I don't think it was accounted for if it did.

    http://www.baskent.edu.tr/~aerdamar/ch2ex.pdf [Broken] Problem 2.89

    I checked my work using this site and I think -20.8 is right, although I'm still not sure. What do you think? Thanks.

    Does part B look good as well?
     
    Last edited by a moderator: May 5, 2017
  18. Sep 13, 2011 #17
    so, if the actual time it took to get down from rest is 1.8 seconds, then Vy(1.8)=-1/2g(1.8)^2= -15.89m/s
     
  19. Sep 13, 2011 #18
    nevermind. thanks
     
  20. Sep 13, 2011 #19
    the only way you can account for the 3.26 seconds automatically is if its in free fall or maybe if its going some initial velocity upwards. ok ok ok yea you are righ

    heres the proof.

    y=v0t - 1/2gt^2

    16=v03.26 - 1/2g 3.26^2

    then v0=20.898

    so, vy(3.26s)=20.898 - 9.81(3.26s)=-11.1m/s
    whew yes you are right
     
  21. Sep 14, 2011 #20
    Thanks for your time! Does part B look good as well?
    I think I understand the constant acceleration equations a little better after this discussion.
     
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