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Dynamics, falling balls.

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data

    [​IMG]

    2. Relevant equations

    I'm not sure what equations to use here, or even wether or not I need to use calc. Unfortunately we just started dynamics and most students don't know how to integrate yet (including me)

    v = dx/dt
    a = dv/dt = d2x/dt2 = v dv/dx


    Uniform rectilinear motion: x = x0 + vt
    Uniformly accelerated rectilinear motion:
    v = vo + at ; x = xo + vot + 1/2at2; v2 = v2o +2a(x-xo)
    where o in all the above = initial.


    3. The attempt at a solution
    I don't even know where to start with this one. hence my question for help.
    We know that a = 9.81m/s2 and also that every .5 second there is a ball dropped.
    Do i have to integrate anything?
     
    Last edited: Nov 21, 2008
  2. jcsd
  3. Nov 20, 2008 #2
    As you have stated already, a is a constant (9.81 m/s^2). With uniform acceleration, there is no need to know how to integrate.
     
  4. Nov 20, 2008 #3
    ok. Well I think I have a solution then, but it's not very fancy (I just used techniques from high school physics).
    1) I first contemplate ball #1 to find the final velocity and the time it is at 3m.
    Vf2=Vi2+2ax
    Vf2= 02+2(9.81)(3)
    Vf=7.67m/s
    Now the time:
    x=( (Vf+Vi)/2)t ; 3m=((7.67+0)/2)t ; t= .78206s

    Now I say Ball #2 is dropped .5s after ball #1 and solve for it's final velocity
    a=(Vf-Vi)/t ; Vf= 2.767m/s

    Now I solve for displacement between ball 1 and 2 according to their respective instantanous(or final) velocities and time being dropped apart from each other.
    Where x is displacement.
    x=( (Vf+Vi)/2)t
    x= ((7.67+2.767)/2)*0.5
    x= 2.6097m

    Can someone confirm this is correct methodology and answer?
    Thanks,
     
  5. Nov 20, 2008 #4
    High school physics is all I use everyday :smile:

    This part is correct. (It was probably easier to use the formula for displacement, x = Vi t + 1/2 a t2.)

    I don't quite understand your explanation for this part of your solution, but if you're saying what I think you're saying, the answer is correct! Are you saying that the problem is equivalent to the distance moved by a ball starting with an initial velocity of 2.767m/s in 0.5 sec?

    A simpler solution might be to just find what distance ball #2 travels in 0.28206 seconds.
     
  6. Nov 20, 2008 #5
    Actually.. i didn't articulate that very well i suppose. But you're right, if I just solved for the distance ball #2 travelled and deducted that figure from 3m it should tell me the spread between them and hence h (or x).
     
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