Page:Somerville Mechanism of the heavens.djvu/121

or, eliminating $dt$ by means of the preceding integral, and making

$-{\frac {g}{2C^{2}}}=a$,

it becomes

${\frac {du}{dx}}=2ac^{2h}$

The integral of this equation will give $u$ in functions of $x$, and when substituted in

$dz=udx$,

it will furnish a new equation of the first order between $z,x,$ and $t$, which will be the differential equation of the trajectory.

If the resistance of the medium be zero, $h=0$, and the preceding equation gives

$u=2ax+b,$

and substituting ${\frac {dz}{dx}}$ for $u$, and integrating again

$z=ax^{2}+bx+b'$

$b$ and $b'$ being arbitrary constant quantities. This is the equation to a parabola whose axis is vertical, which is the curve a projectile would describe in vacuo. When

$h=0,d^{2}z=gdt^{2};$

and as the second differential of the preceding integral gives

$d^{2}z=2adx^{2};dt=dx{\sqrt {\frac {2a}{g}}}$,

therefore

$t=x{\sqrt {\frac {2a}{g}}}+a'$.

If the particle begins to move from the origin of the co-ordinates, the time as well as $x,y,z$, are estimated from that point; hence $b'$ and $a'$ are zero, and the two equations of motion become

$z=ax^{2}+bx$; and $t=x{\sqrt {\frac {2a}{g}}}$;

whence

$z=g{\frac {t^{2}}{2}}+tb{\sqrt {\frac {g}{2a}}}$.