Question 9.5: Solve the equations of motion for a projectile in space by t......

With the z-axis oriented vertically upwards,

H =\frac{1}{2m}\left(p^{2}_{x}+p^{2}_{y}+p^{2}_{z}\right)+mgz,   (9.70)

and the Hamilton-Jacobi equation is

\frac{1}{2m}\left[\left(\frac{\partial S}{\partial x } \right)^{2} +\left(\frac{\partial S}{\partial y} \right)^{2}+\left(\frac{\partial S}{\partial z } \right)^{2}\right]+mgz+\frac{\partial S}{\partial t }=0.  (9.71)

Since x and y are cyclic coordinates and H does not explicitly depend on time,

S = −\alpha _{1}t + \alpha _{x} x+ \alpha _{y}y + W(z) , (9.72)

whence

\left(\frac{dW}{dz}\right)^{2}=2m\left(\alpha _{1}-mgz\right)-\alpha^{2} _{x}-\alpha^{2} _{y}.  (9.73)

Integrating this equation we find

S =−\alpha _{1}t + \alpha _{x} x+ \alpha _{y}y- \frac{1}{3m^{2}g}\left[2m\left(\alpha _{1}-mgz\right)-\alpha^{2} _{x}-\alpha^{2} _{y}\right] ^{{3}/{2}}.  (9.74)

The motion of the particle is determined by means of

\beta_{1}=\frac{\partial S}{\partial \alpha _{1} }=-t-\frac{1}{mg}\left[2m\left(\alpha _{1}-mgz\right)-\alpha^{2} _{x}-\alpha^{2} _{y}\right]^{{1}/{2}},  (9.75a)

\beta_{2}=\frac{\partial S}{\partial \alpha _{x} }=x+\frac{\alpha _{x}}{m^{2}g}\left[2m\left(\alpha _{1}-mgz\right)-\alpha^{2} _{x}-\alpha^{2} _{y}\right]^{{1}/{2}},  (9.75b)

\beta_{3}=\frac{\partial S}{\partial \alpha _{y} }=y+\frac{\alpha _{y}}{m^{2}g}\left[2m\left(\alpha _{1}-mgz\right)-\alpha^{2} _{x}-\alpha^{2} _{y}\right]^{{1}/{2}}.  (9.75c)

The resolution of these three last equations for x, y, z yields

x = A + \frac{\alpha _{x}}{m}t,  (9.76a)

y= B+ \frac{\alpha _{y}}{m}t,  (9.76b)

z = C + Dt- \frac{gt^{2}}{2},  (9.76c)

where the constants A, B,C,D are given in terms of the αs and βs. Equations (9.76) coincide with the usual solution of this problem by elementary means.