Page:Somerville Mechanism of the heavens.djvu/121

Chap II.] 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^2z = gdt^2; $

and as the second differential of the preceding integral gives

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

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

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

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