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Two Stones Falling

  1. Jun 26, 2003 #1
    Ok, so there are two stones. One is dropped (not thrown) off a building. 2 seconds later a rock is thrown at 30 m/s. The rocks hit the ground at the same time. Here's what I have so far:

    R1:

    v0 = 0 m/s
    T = unknown
    x = o m
    x0 = unknown
    a = 9.8 m/s^2

    R2:

    v0 = 30 m/s
    T = Time of R1 - 2 s
    x = o m
    x0 = unknown
    a = 30 m/s @ 9.80 m/s^2

    I know that these two equations must equal eachother:

    (.5)(9.80 m/s^2)(T)^s = (30m/s)(T-2) + (.5)(9.80m/s^2)(T-2)^2

    However, I have expanded this a million times and for some reason I can't get things to click. I mean, I could do guess and check but there has got to be an easier way...

    The question asks how long it took the first stone to reach the gound, how high the building is, and what speeds the stones are at before they hit the ground.

    Since they ask for the time first, I thought I would have to use the 2nd kinematic equation. But then I realized that I dont know what the initial height was. I feel stuck because I have these two variables the initial height and time. What do I use to find these one at a time? Lol, I'm stuck.
     
  2. jcsd
  3. Jun 26, 2003 #2

    Tom Mattson

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    There is an easy way: The quadratic formula.

    Expand out the square, and you'll see that you have a simple quadratic polynomial. You can then solve by:

    x=(-b±(b2-4ac)1/2)/2a
     
  4. Jun 26, 2003 #3
    Comes down to simplfied before plugging into equation:

    (-19.6 m/s^2)T + (30 m/s)T + (19.6 m/s^2) - (60 m/s)

    How does this follow std form?
     
  5. Jun 26, 2003 #4

    Tom Mattson

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    You messed up somewhere, because you should have a "T2" term in there. Slow down and have another crack at it.
     
  6. Jun 27, 2003 #5

    HallsofIvy

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    You clearly mean T2 in the first term because you will need seconds squared to cancel the s^2 term in -19.6 m/s^2.

    For the same reason "(19.6 m/s^2)- (60 m/s)" is impossible- you can't add (or subtract) velocity and acceleration.

    Finally, to solve an equation, you have to have an equation! What is this supposed to be equal to?
     
  7. Jun 27, 2003 #6
    Ok, I was looking at my physics book a little more...

    You can find time with the quadratic equation with this equation...

    y = y0 + v0t + .5at^2

    However, you cannot solve for time with this method unless you have all the information except time. In my case, this is what I have for the second stone:

    x = 0
    x0 = unknown
    t = unknown
    a = 9.80 m/s
    v0 = 30 m/s (it was thrown not dropped)

    Due the fact that there are two unknowns I cannot solve with the quadratic. The other stone, which was dropped, must be included in the mix somewhere - it has to be a two part problem that I can't see.

    First stone:

    x = 0
    x0 = unknown
    t = unknown
    a = 9.0 m/s^2
    v0 = 0 m/s

    If you think about it these two equations must equal eachother if the two rocks hit the ground at the same time:

    (.5)(9.80 m/s^2)(T)^2 = (30m/s)(T-2) + (.5)(9.80m/s^2)(T-2)^2

    I have the actual answer, its in the back of the book, and if you plug the time into it, the equations equal eachother. However, I have expanded this equation a million times. the (4.90 m/s^2)T^2 always cancels out (which doesnt give rise to std form for the quadratic equation. I'm really stuck guys and gals.
     
  8. Jun 27, 2003 #7

    HallsofIvy

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    Okay, your equation is:

    4.9T2= 4.9(T-2)2+ 30(T-2)

    multipling that out:

    4.9T2= 4.9(T2- 4T+ 4)+ 30T- 60
    = 4.9T2- 19.6T+ 19.6+ 30T- 60

    The "4.9T2" terms on both sides cancel! We don't have a quadratic equation after all!

    We do have, after we move the terms involving T to one side,
    we have 30T- 19.6T= 60- 19.6 or 10.4T= 40.4 so T= 40.4/10.4= 3.88 seconds. Now, you can use that value of T to find the height of the building.
     
  9. Jun 27, 2003 #8
    Is it ok that you neglected units like that? Like m/s^s and m/s and s?
     
  10. Jun 27, 2003 #9
    Well, I guess you can, because according to the book that answer is correct. It was easy to find the height based off of this, but what does "what speeds the stones are at before they hit the ground" mean? "Just before" could be ay any point couldn't it?
     
  11. Jun 27, 2003 #10
    Oh wait I understand it now, when they say right before they mean the maximum velocity of the first and second stones. That is a stupid way for the book to word it, lol.
     
  12. Jun 29, 2003 #11

    HallsofIvy

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    First, I didn't need to carry the units along because I had checked them and knew they were correct.

    Second, any time a text book seems to word something in a "stupid way" look again- you might learn something. (I'm not saying it CAN'T be stupid- just that authors (and editors) tend to be very careful about wording.)

    In this case, if they had said "maximum velocity" that would have been assuming things not yet given. The point of "just before" (I notice you say "before they hit the ground" and "just before"- which was it?) is that you can use the time you have already calculated for hitting the ground. AT the time they hit the ground they start decellerating rapidly so that would be ambiguous.
     
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