A child's swing of length L, attached y = L + y0 above the ground, reaches a maximum heigh of y = h during the swing.
I am trying to write vector equations for the instantaneous velocity, as a function of L, g, h, y and y0.
I am also trying to write an equation for the time it will take after leaving the swing until landing on the ground; again as a function of: v(x), v(y), g and y0.
Then I am trying to write an equation for the horizontal distance travelled as a function of L, g, h, y and y0.
Finally, I am trying to provide a simplified equation for the horizontal distance travelled as function of L, g, h, and y, when y0 = 0.
Thank you to anyone who can help me.

I suggest you start with the equation of motion for the swing in polar coordinates, which for small angles of displacement is:


\ddot \theta - \frac g L \theta =0


The solution of this differential equation is:


\theta = A \sin (\sqrt{\frac g L} t)


where A = the max displacement of the swing in radians. The velocity equation is the derivative of this:


\dot \theta = A \sqrt{ \frac g L} \cos (\sqrt {\frac g L } t)


To get this in terms of angle theta replace the cosine term with a sine equivalent:


\cos(\sqrt { \frac g L } t) = \sqrt { 1- \sin^2(\sqrt { \frac g L} t) } = \sqrt {1 - \frac {\theta^2}{A^2} }

Hence:

\dot \theta = \sqrt {\frac g L} \sqrt {A^2 - \theta^2}

Now to get into terms of linear velocity rather than radial velocity:


v = L \dot \theta = \sqrt {gL}\sqrt {A^2 - \theta^2}


The vector components of v in cartesian coordinates can then be found using:


v_x = v \cos\theta \\
v_y = v \sin\theta


So now you have x and y components of velocity for any angle \theta of the swing. When you jump off the swing at angle \theta you can calculate distance and altitide achieved using standard equations of motion for a projectile. You will have to convert \theta into values of h to get it into the terms you want, but that's fairly trivial.

Hope this helps.

I suggest you start with the equation of motion for the swing in polar coordinates, which for small angles of displacement is:


\ddot \theta - \frac g L \theta =0


The solution of this differential equation is:


\theta = A \sin (\sqrt{\frac g L} t)


where A = the max displacement of the swing in radians. The velocity equation is the derivative of this:


\dot \theta = A \sqrt{ \frac g L} \cos (\sqrt {\frac g L } t)


To get this in terms of angle theta replace the cosine term with a sine equivalent:


\cos(\sqrt { \frac g L } t) = \sqrt { 1- \sin^2(\sqrt { \frac g L} t) } = \sqrt {1 - \frac {\theta^2}{A^2} }

Hence:

\dot \theta = \sqrt {\frac g L} \sqrt {A^2 - \theta^2}

Now to get into terms of linear velocity rather than radial velocity:


v = L \dot \theta = \sqrt {gL}\sqrt {A^2 - \theta^2}


The vector components of v in cartesian coordinates can then be found using:


v_x = v \cos\theta \\
v_y = v \sin\theta


So now you have x and y components of velocity for any angle \theta of the swing. When you jump off the swing at angle \theta you can calculate distance and altitide achieved using standard equations of motion for a projectile. You will have to convert \theta into values of h to get it into the terms you want, but that's fairly trivial.

Hope this helps.

Thank you for the very helpful pointer towards solving this problem.
Forgive me for asking, but how would convert theta into values of h to get it into the necessary terms?

Forgive me for asking, but how would convert theta into values of h to get it into the necessary terms?

h = y_0+L(1-\sin \theta)

h = y_0+L(1-\sin \theta)

Sorry, but should this be:
h = y0 + L (1 - cos theta) ? -
because the maximum height that the swing could reach the grind would be
h = y0 + ( L - L cos theta).

Also, is the angle theta that the swing makes with the vertical (as described above) not different from the angle that the child launches himself from the swing?
So how is it possible to use this angle in the equations for projectile motion when one is attempting to find the horizontal distance that the child lands from the swing as a function of L, g, h, y and y0

sheila - good catch. I agree that it should be


h = y_0 + L(1 - \cos \theta)


The angle of launch of the chid from the horizontal is equal to \theta.

sheila - good catch. I agree that it should be


h = y_0 + L(1 - \cos \theta)


The angle of launch of the chid from the horizontal is equal to \theta.

Oh yes, I see now. Thanks for that. The picture helps.