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Rocks falling at the same time and their distance apart

  1. Oct 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Two rocks are falling from a roof. The second one begins its fall when the first one has already covered 30cm. Calculate the distance d between these two rocks after 10 seconds counted from the moment the second rock begins its fall.

    2. Relevant equations

    [itex]s=s_{0}+v_{0}t+\frac{at^{2}}{2}[/itex]
    [itex]v(t) = v_0 + at[/itex]

    3. The attempt at a solution

    It turns out second rock will go as follows: [itex]s_2=\frac{at^2}{2}={9,81*100}{2}\approx 490,5m[/itex]. If it does, then we have [itex]0,3=\frac{9,81t^2}{2}[/itex] so [itex]t\approx0,25s[/itex]. Then, we know that [itex]v(0,25)=0+9,81*0,25=2,4525m/s[/itex]. And finally [itex]s_1=0,3m+2,45*10+\frac{9,81\cdot100}{2}=515,3m[/itex] so they're 24,8 metres apart. Is that OK?
     
  2. jcsd
  3. Oct 15, 2012 #2
    Is something wrong here? I can't find any but something surely does not feel right.
     
  4. Oct 15, 2012 #3
    What seems wrong?
     
  5. Oct 15, 2012 #4
    I got to the same result.
    Basically you compare distances.
    s1 = Δs + g(0.247+10)²/2
    s2 = 0 + g10²/2

    s1 - s2 = Δs + g/2(10.247²-10²) = 0.3 + 24.529 ~ 24.8
    Δs is the freefall during the 1st 0.247 or 0.25 seconds - the 30cm given.

    I am fairly certain you have done it correctly.
     
  6. Oct 15, 2012 #5
    Oh... That's good. Thank you :) Something felt wrong as I'm simply not too confident with my physics skills ;)
     
  7. Oct 15, 2012 #6
    I'm having trouble understanding the first line - s1 = Δs + g(0.247+10)²/2. Isn't Δs already taken into account with these 0.247 sec?
     
  8. Oct 15, 2012 #7
    Can't find a comprehendable explanation - my lingual skills fail me :(

    The way I understand it is that if I plug 10s into both of these equations, they both cancel out and the constant distance between them is always 30cm - but that makes no sense, because the 1st stone already has a base velocity of 2.something metres per second and the only way to express it in the equation is to use the time on top of the 10s to show that s1 doesn't start at the same speed as s2.
     
    Last edited: Oct 15, 2012
  9. Oct 15, 2012 #8
    Please take into consideration following:

    10.25^2*9.81*0.5-10^2*9.81*0.5=24.83.

    10.247^2*9.81*0.5-10^2*9.81*0.5=24.52.
     
  10. Oct 15, 2012 #9
    Since the acceleration is the same in both cases, the only difference in distances will be due to the initial difference (30 cm) and the initial velocity of the first rock (2.45 m/s according to your calculation). Thus diff = 30 cm + 2.45 m/s * 10 s = 24.8 m.
     
  11. Oct 15, 2012 #10

    HallsofIvy

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    Here's how I would have done the problem: Since the first rock is dropped its initial speed is 0 and so the distance dropped is [itex]d_1= (g/2)t^2[/itex]. It will have dropped 30 m when [itex](g/2)t^2= 30[/itex] so [itex]t= \sqrt{60/g}[/itex].

    The second rock is dropped at that time also with initial speed 0 and so [itex]d_2= (g/2)(t- \sqrt{60/g})^2= (g/2)t^2- g\sqrt{60/g}t+ 30[/itex].

    Subtracting, the distance between the two rocks, after t seconds, is [itex]g\sqrt{60/g}t- 30[/itex].
     
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