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Continuity Conditions

  1. Feb 27, 2012 #1
    The problem statement, all variables and given/known data

    A ball falls from rest at a height H above a lake. Let y = 0 at the surface of the lake. As the ball falls, it experiences a gravitational force -mg. When it enters the water, it experiences a buoyant force B so the net force in the water is B - mg.

    a) Write an expression for v(t) and y(t) while the ball is falling in air.
    b) In the water, let v2(t) = at + b and y2(t) = (1/2)at2 + bt + c where a = (B - mg)/m. Use
    continuity conditions at the surface of the water to find the constants b and c.

    The attempt at a solution

    a)
    Since [itex]\vec{v}=\int\vec{a}dt=\vec{a}t+\vec{v}_{0}[/itex] and the initial velocity is 0, we have [itex]\vec{v}(t)=\vec{a}t[/itex]. Using [itex]\vec{F}=m\vec{a}[/itex] yields [itex]v(t)=-gt[/itex].
    Also, [itex]\vec{r}=\int\vec{v}dt=\frac{1}{2}\vec{a}t^{2}+\vec{v}t+\vec{r}_{0}[/itex]. Again, since the initial velocity is 0 and [itex]y=r-r_{0}[/itex] we have [itex]y(t)=-\frac{1}{2}gt^{2}[/itex].
     
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  3. Feb 27, 2012 #2

    SammyS

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    Isn't the ball at y = H, at time t=0 ?
     
  4. Feb 27, 2012 #3
    Yes. So the last equation should be [itex]y(t)=-\frac{1}{2}gt^{2} + H[/itex]. Can you help with b)? I do not understand the question.
     
  5. Feb 27, 2012 #4

    SammyS

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    At what time does the ball enter the water?
     
  6. Feb 27, 2012 #5
    When y = 0, [itex]-\frac{1}{2}gt^{2}+H=0[/itex] and so [itex]t=\sqrt{\frac{2H}{g}}[/itex].
     
  7. Feb 28, 2012 #6
    I do not understand the question. Please help.
     
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