User:Evil4Zerggin/Suborbital trajectory

It appears that gravity drops linearly from its maximum value of 9.81 m/s (henceforce referred to as $$g_0$$) at 500 m altitude to zero at 900 m. Let $$y_g = y - 900$$, or the altitude relative to the edge of gravity. In this range, we have

$$g = \frac{d^2 y_g}{dt^2} = g_0 \frac{y_g}{400}$$

This is similar to the equation of a harmonic oscillator, except the sign is the opposite.

The solution to this equation is of the form

$$y_g \left(t \right) = a e^{0.05 \sqrt{g} t} + b e^{-0.05 \sqrt{g} t}$$

To find $$a, b$$, we use our initial conditions $$y_{g0}, v_0$$:

$$a + b = y_{g0}$$

$$a - b = \frac{20 v_0}{\sqrt{g}}$$

whence

$$a = \frac{y_{g0} + \frac{20 v_0}{\sqrt{g}}}{2}$$

$$b = \frac{y_{g0} - \frac{20 v_0}{\sqrt{g}}}{2}$$

In analogy to the solution to the harmonic oscillator being a sum of sine and cosine, we can also express this as a sum of hyperbolic sine and cosine:

$$y_g \left(t \right) = y_{g0} \cosh \left( 0.05 \sqrt{g} t \right) + \frac{20 v_0}{\sqrt{g}} \sinh \left( 0.05 \sqrt{g} t \right)$$