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Not sure what is being asked. motion i sure but i dont really understand question

  1. Jan 30, 2013 #1

    veg

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    1. The problem statement, all variables and given/known data

    Your roommate accidentally tosses your Frisbee onto the roof of a 16m high building. You figure you are going to have to go looking for a ladder, but as you stare up at your favorite competition disk, you realize you could possibly knock it loose with a stone. You don’t want to look like an idiot in front of your roommate however, and the well in the courtyard suddenly gives you an idea. You drop a stone into the well and note that you hear a splash 2.0s later. You repeat the experiment with another stone, but this time you throw the stone down as fast as you can. This time the splash comes 1.0s after the stone leaves your hand. Armed with this information you carry out a quick calculation. Can you use the stone, or do you need to go get a ladder?

    2. Relevant equations

    so far i used this equation to find Vf or final velocity which im sure is height of Δx
    Vf = Vi + aΔt
    Vf = 0m/s + 10m/s (2s) = 20

    3. The attempt at a solution

    so from what i can kind of get from this problem is that there is 3 different scenarios i need to solve for. i get the first scenario or easy one. im just not sure what to do about the other two or if i am on the right track?
     
  2. jcsd
  3. Jan 30, 2013 #2

    jedishrfu

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    Welcome to PF.

    You need to show us your work and we can give you hints on how to solve the problem. We can't do the work for you.

    So basically you need to decide: use a ladder ot throw a stone

    What does dropping a stone in the well tell you about the well?

    Throwing a stone down the well what does that tell you about your throwing capablity?

    From what you now know which is it use a ladder or throw a stone?
     
  4. Jan 30, 2013 #3

    rude man

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    I gave away too much.
     
  5. Jan 30, 2013 #4

    veg

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    Well i drew three pictures. the first picture is a well with initial velocity of 0m/s. an acceleration of 10m/s (problem said use 10 not 9.8) a time of 2 seconds. which is basically a simple free fall question so...

    Vf = Vi + a * t
    Vf = 0m/s + 10m/s * 2 seconds

    dropping the stone gives final velocity which would also be height of well because it falls at constant speed so the well is 20 meters deep (delta x).

    since the well is 20 meters deep i would assume we could hit the frisbee since it is only
    16m which is 4 meters less than the 20 meter well.

    but im not sure how to apply this to solve what i see as two other problems
    which would be the initial velocity of the stone in the second well and not even sure what to do about the 16 m high building.
     
  6. Jan 30, 2013 #5

    rude man

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    Knowing the depth of the well and the time it took a rock to hit the bottom, you can compute the initial velocity with which the rock was thrown down the well. BTW it's not 20m, it's somewhat less than that.

    You cannot compare 16m with the well depth directly.
     
  7. Jan 30, 2013 #6

    veg

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    so based on what you said i would take this equation Vf = Vi +at and rearrange it to get this Vi = Vf - at so plugging in numbers i would get Vi = 20m/s - 10m/s * 1s = 10

    also not sure what you mean by "it's not 20m, its somewhat less than that" do you mean im off. Only thing i could think of was using 9.8m/s for a instead of 10m/s but problem said to use 10m/s
     
  8. Jan 30, 2013 #7

    haruspex

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    You appear to be confused between acceleration, speed and distance.
    The stone falls, whether thrown or dropped, at constant acceleration. You computed the final speed when dropped. You have not computed the depth of the well.
     
  9. Jan 30, 2013 #8

    veg

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    Δx = Vi* t + .5at^2

    Δx = 0m/s * 2 + .5(10)2^2

    Δx = 0 + 20

    so it would just be 20 meters deep right?
     
  10. Jan 30, 2013 #9

    rude man

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    No, your equation should be s = vi*t + at^2/2 where s is the depth you just calculated.

    If you're using g = 10m/sec^2 then s = 20m is right. I think your instructor was dumb to have you use g = 10 m/sec^2.
     
  11. Jan 30, 2013 #10

    veg

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    yeah, when i read that i wasn't sure why 10 was being used. ill have to ask because its usually 9.8. but now that i know depth of well (20m) for the first part of problem then the initial velocity would be 10m/s for the second part of problem (throwing rock in well right??). This is where i am now stuck. I am not sure how to apply this to getting a ladder or throwing a rock. so here is what i know and am confused about

    a = 10
    x = 20
    Vi = not sure one is 0m/s for free fall and the other is 10m/s for throwing it in well
    Vf = 20m/s
    t = not sure two are given 1 and 2 seconds for two problems
     
  12. Jan 30, 2013 #11

    rude man

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    Never mind vf. Use s = vi*t + at^2/2 to get vi.

    You needed the 2 sec to determine the well depth. Now you need the 1 sec to determine vi.
     
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