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Motion Problem

  1. Oct 22, 2006 #1
    Hi, I have a kinematics problem involving free fall acceleration. I'm not sure how to do this so I was wondering if someone could please give me a hand. Thank you.

    Question: Water drips from the nozzle of a shower onto the floor [tex]2m[/tex] below. The drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. When the first drop strikes the floor, how far belowthe nozzle are the second drop?

    I started the problem off by figuring out how long it would take for the first drop to hit the floor: [tex]-2.00m=-\frac{1}{2}(9.8m/s^2)t^2[/tex]
    The time was [tex]t=\sqrt{\frac{2.00m}{4.9m/s^2}}\approx 0.64s[/tex]

    Now how can I use this information to solve this problem?
     
    Last edited: Oct 22, 2006
  2. jcsd
  3. Oct 22, 2006 #2

    Hootenanny

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    Firstly, your time is not correct. You have an error in your manipulation; the correct expression is as follows;

    [tex]x=\frac{1}{2}at^2 \Leftrightarrow t = \sqrt{\frac{2x}{a}}[/tex]

    You know that within the time take for the first drop to hit the floor, the fourth drop has just been produced. What is the time interval between successive drops?
     
  4. Oct 22, 2006 #3

    Office_Shredder

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    Hoot, the time looks OK to me. 2*2/9.8 = 2/4.9

    He just divided the 9.8 by 2 before bringing it to the other side.
     
  5. Oct 22, 2006 #4
    So should I find the how many seconds for one drop to be produce, i.e. [tex]\sqrt{\frac{2x}{g}}\times\frac{1}{4}[/tex]

    Then to find the position of the second drop plug into [tex]x=\frac{1}{2}gt^2[/tex], [tex]t=\frac{3}{4}\sqrt{\frac{2}{4.9}}s[/tex]?
     
  6. Oct 22, 2006 #5

    Hootenanny

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    Indeed it is, my apologies! I think its time for a break :zzz: .
     
  7. Oct 22, 2006 #6
    Let t=.64 sec=T

    the first drop is released at t=0 sec, the last at t=T
    since it is regular, when are the second and third drops released?
     
  8. Oct 22, 2006 #7

    Hootenanny

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    Yes, so you plug

    [tex]t=\frac{2}{3}\sqrt{\frac{2}{4.9}}s[/tex]

    into

    [tex]x=\frac{1}{2}gt^2[/tex]

    To find the distance traveled.
     
    Last edited: Oct 22, 2006
  9. Oct 22, 2006 #8
    Okay, then [tex]t=\frac{3}{4}\sqrt{\frac{2}{4.9}}s\approx 0.48s[/tex]
    Plugging into, [tex]x=\frac{1}{2}gt^2=\frac{1}{2}(9.8m/s^2)(.48s)=1.125m[/tex]

    Is this correct?
     
  10. Oct 22, 2006 #9
    No...
    I've done it wrong, I think. I should be multiplying [tex]\frac{2}{3}[/tex] to the time instead of [tex]\frac{3}{4}[/tex] otherwise, it would mean once a drop starts to drip, the 4th drop below it had hit the ground 2.00 m below.
     
  11. Oct 22, 2006 #10
    2/3 is correct since it starts to fall at t=1/3 (.64)
    so now it's not too hard.
     
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