- #1

Bazzinga

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I'm having trouble with part a) of this question...

[PLAIN]http://img69.imageshack.us/img69/5815/98157006.png

So I started off by solving the DE above a), and I've gotten it down to:

[tex]\frac{1}{2} m v^{2} = \frac{mgR^{2}}{(x + R)} + C[/tex]

I can tell I'm getting close, but I'm a little confused where h is coming from, and how to get rid of the [tex]R^{2}[/tex]. Also, what do I do with C??

I know it says x = x(t) which is the height, so does that mean that x(t) = h? I've never taken physics (and this was in my Calculus homework) so this is all new to me!

[PLAIN]http://img69.imageshack.us/img69/5815/98157006.png

So I started off by solving the DE above a), and I've gotten it down to:

[tex]\frac{1}{2} m v^{2} = \frac{mgR^{2}}{(x + R)} + C[/tex]

I can tell I'm getting close, but I'm a little confused where h is coming from, and how to get rid of the [tex]R^{2}[/tex]. Also, what do I do with C??

I know it says x = x(t) which is the height, so does that mean that x(t) = h? I've never taken physics (and this was in my Calculus homework) so this is all new to me!

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