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All I could think of was to write the acceleration as a function of height and solve the new equation for height. Is this allowed? Below is what I put together. Obviously ignoring air resistance for now. Any comments/suggestions would be helpful. Am I on the right track?

Starting with $$s=vt-\frac{1}{2}at^{2}$$

Where ##s## is the height, ##v##, final velocity, ##t##, time of flight and ##a## the acceleration

Then changing ##a=-g## to $$a=-\frac{Gm_{\oplus }}{\left ( r_{\oplus }+s \right )^{2}}$$

Where ##m_{\oplus }## and ##r_{\oplus }## are respectively the mass and radius of the Earth

At maximum height ##v=0##, so the equation becomes $$s=\frac{Gm_{\oplus }t^{2}}{2\left ( r_{\oplus }+s \right )^{2}}$$

Rearranging gives a cubic in ##s##

$$2s^{3}+4r_{\oplus }s^{2}+2r_{\oplus }^{2}s-Gm_{\oplus }t^{2}=0$$

Then subbing in values of the constants ##r_{\oplus }=6.378\times 10^{6}\mathrm{m}##, ##m_{\oplus }=59.736\times 10^{23}\mathrm{kg}##, ##G=6.673\times 10^{-11}\mathrm{m^{3}kg^{-1}s^{-2}}## and ##t=50\mathrm{s}## (assuming symmetrical time of flight about the max height).

$$2s^{3}+2.5512\times 10^{7}s^{2}+8.1357768\times 10^{13}s-3.98618328\times 10^{18}=0$$

Solving (calculator) gives ##s=12.202\mathrm{km}## plus 2 complex roots. The real value is less than when ignoring variable gravitational field strength. Error or explanation? It doesn't make intuitive sense to me at all. It should be more.