- #36

#### wrobel

Science Advisor

- 1,104

- 960

Ok then :) let's develop the variable mass version of the story.

Equation from #11 takes the form

$$-2\frac{d}{dt}(\dot y y)=gy.$$ After a change ##y^2=z## we have

$$\ddot z=-g\sqrt z,$$

and let ##\dot z=u(z)##

then

$$\frac{1}{2}\frac{d}{dz}u^2=-g\sqrt z.$$

Eventually it follows that

$$\dot y=-\frac{1}{y}\sqrt{-\frac{1}{3}g\Big(y^3-\Big(\frac{L}{2}\Big)^3\Big)}.$$

it is clear ##\dot y\to-\infty## as

##y\to 0+##

Equation from #11 takes the form

$$-2\frac{d}{dt}(\dot y y)=gy.$$ After a change ##y^2=z## we have

$$\ddot z=-g\sqrt z,$$

and let ##\dot z=u(z)##

then

$$\frac{1}{2}\frac{d}{dz}u^2=-g\sqrt z.$$

Eventually it follows that

$$\dot y=-\frac{1}{y}\sqrt{-\frac{1}{3}g\Big(y^3-\Big(\frac{L}{2}\Big)^3\Big)}.$$

it is clear ##\dot y\to-\infty## as

##y\to 0+##

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