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Differential Equations

  1. May 19, 2005 #1
    1) Show that since a body falling freely obeys the differential equation [tex]h''=-g[/itex], if it falls from an initial height [itex]h(0)[/itex], it lands with a velocity of [itex]-\sqrt{2gh(0)}[/itex]

    This problem is from a differential equations class and I solved it two different ways:

    Method 1:
    [itex]\frac {dh} {dt}=v[/itex] and [itex]\frac {dv} {dt}=a[/itex] which leads to [itex]\int{a dh}=\int{v dv}[/itex] and solving for v gives you [itex]v=\sqrt{2gh}[/itex]

    Method 2:
    Use [itex]KE=PE[/itex], so [itex]\frac{1}{2}mv^2=mgh[/itex]
    and solve for v which gives you [itex]v=\sqrt{2gh}[/itex].

    Unfortunately I don't think this is what the instructor is looking for. The wording of the question suggests he wants me to use differential equations techniques. I solved the DE to get [itex] h=-\frac{1}{2}gt^2+c_1t+c_2 [/itex] but I'm not seeing how that will get me to [itex]v=-\sqrt{2gh(0)}[/itex]

    Can anyone help me?
  2. jcsd
  3. May 19, 2005 #2


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    The velocity's equation v(t) is [itex] v(t)=v_{0}-gt [/itex] If u take the initial velocity to be zero and u found the falling time in terms of the height,then it's easy to get

    [tex]v\left(t_{\mbox{falling}\right)=-\sqrt{2g h(0)} [/tex]

  4. May 20, 2005 #3


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    Actually, your "method 1" is a differential equation method.

    However, you can, as you say get [tex]h(t)= -\frac{1}{2}gt^2+ c_1t+ c_2[/tex] and then [tex]v(t)= h'(t)= -gt+ c_1[/tex].
    Taking t= 0 to be the moment the body is dropped, h(0)= c_2 and v(0)= c_1= 0 so
    [tex]h(t)= -\frac{1}{2}gt^2+ h(0)[/tex] and [tex]v(t)= -gt[/tex].

    The body "lands" when h(t)= 0. Solve [tex]h(t)= -\frac{1}{2}gt^2+ h(0)= 0[/tex], which is thesame as [tex]t^2= \frac{2h(0)}{g}[/tex], (of course, you only want the positive root) and put into v(t)= -gt to find speed with which it lands.
  5. May 21, 2005 #4
    Thanks for helping me out...I don't know why, but the answer wasn't immediately apparent to me. I need a break :tongue:
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