# 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:

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