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Falling objects puzzle

  1. Jul 28, 2004 #1
    two dimensions

    Two stones are thrown horizontally with the same velocity from the tops of two different buildings. One stone lands twice as far from the based of the building from which it was thrown as does the other stone. Find the ratiio of the height of the taller building to the height of the shorter building.
  2. jcsd
  3. Jul 29, 2004 #2
    Horisontal velocity is constant.
    distance = velocity * time
    As the second stone travel twice the distance of the first one, it must bee falling for twice the time:

    Hope that helped.
  4. Jul 29, 2004 #3

    If it falls twice the distance and it takes twice the time, then that means it's twice as high, right? The ratio would be 2 to 1? That seems logical but it seems too simple to be correct. What am I not seeing?
  5. Jul 29, 2004 #4

    Doc Al

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    Staff: Mentor

    No one said it fell (vertically) twice the distance! It fell for twice the time. Now you have to know how distance fallen relates to time. The distance fallen is proportional to the square of the time: [itex]d = 1/2gt^2[/itex]

    So... if the time to fall is double, what can you say about the relative height of the buildings?
  6. Jul 29, 2004 #5
    this question looks tricky but if you know how to tackle it its really quite easy. Doc Al said it right. If you do the calculation, you would get this:
    Time of the taller building = 2 times the time of the smaller building, since horizontal distance is double
    Horizontal Distance travelled by the projectile of the taller building = 2 times smaller

    Now we can work with the verticla component and time in order to get the height:
    Shorter Building: d = vot + 1/2at^2
    d = 1/2at^2
    t = sqrt(2d/a)

    Taller Building: d = vot + 1/2at^2
    d = 1/2a(2t)^2
    t = sqrt(8d/a)
    Therefore, the ratio of building heights is 4:1.
  7. Jul 29, 2004 #6
    Most questions like these end up with squares.
    I use to think that since the vertical distance is a quadratic function of time, "2 times the time" must mean 2^2 times the distance. If the stone from the taller building flew three times the distance of the other stone, height ratio would be 3^2=9 to 1.

    Assuming that we are neglecting atmospheric drag... =)

    Last edited: Jul 29, 2004
  8. Aug 5, 2004 #7
    Isnt this a bogus question to start with? Seeing we cant measure time properly. Could it even possible for those two stones to land at a precise moment in time simultaneously?

    What would the odds be on that? Can someone give me odds of two objects simultaneously touching the same point since we dont have a proper measurement of time? Would you have to go down to the quantum level to even start to guestimate?

    Just thinking here, nm if you dont want to continue.
  9. Aug 7, 2004 #8


    Using parametric equations gives a simple solution

    y1(t) = h - at^2/2
    y2(t) = H - at^2/2

    solve yn(t) = 0

    t1 = sqrt(2h/a)
    t2 = sqrt(2H/a)

    x1(t) = vo*t
    x2(t) = vo*t

    r1 = vo*t1
    r2 = vo*t2

    r2/r1 = (vo*t2)/(vo*t1)

    r2/r1 = t2/t1 and r2 = 2r1

    2 = sqrt(2H/a)/sqrt(2h/a)
    4 = (2H/a)*(a/2h)

    4 = H/h
  10. Aug 8, 2004 #9

    Doc Al

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    Why do you assume that the stones land simultaneously? All we know is that the stones are thrown with the same speed. No one said they are thrown or land at the same time. And it's not relevant anyway.
  11. Aug 8, 2004 #10
    The stones are thrown at the same horizontal speed, but the vertical heights of the buildings are different. The time it takes for the stone to hit the floor has nothing to do with horizontal speed (now we are neglecting air resistance here), but only to do with the height with which it was dropped. Remember, gravity works in the vertical direction and thus only affects the vertical.
  12. Aug 11, 2004 #11
    Forgive me if this is a stupid question, but I just don't see how Nenad got:
    How do you get the "8" in this answer?

    I am getting sqrt(d/2a)=t

    I am sure there is a simple answer to this. Can anyone point it out to me?
  13. Aug 11, 2004 #12

    Doc Al

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    Beats me what Nenad was up to. The problem is straightforward:
    [tex]D_1 = 1/2 a t_1^2[/tex]

    [tex]D_2 = 1/2 a t_2^2[/tex]

    So, since [tex]t_2 = 2t_1[/tex]:
    [tex]D_2 = 1/2 a (2t_1)^2 = 4(1/2 a t_1^2) = 4D_1[/tex]
  14. Aug 11, 2004 #13
    Thanks Doc Al. That is how I worked it out and came to the 4:1 answer.
    I thought I was missing something with that 8 in Nenad's intermediate steps.
  15. Aug 13, 2004 #14
    Now, can you figure out the answer with the corrections necessary to include the general relativity component induced by the differing heights of the buildings? <is evil>
  16. Aug 13, 2004 #15


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    Yes, considering GR but not air resistance it's D2 = 4.000 D1 for even the tallest building (and half tallest-building) on Earth. :wink:
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