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](adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Homework Help: Differential Equations

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