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I am trying to find expression for h(t) from the equation: $$h''=a(h')^2+g$$ also written as: $$\frac{d^2h}{dt^2}=a\left(\frac{dh}{dt}\right)^2+g$$ I'm teaching myself differential equations and I was applying them to a physics project. I came up with this equation, but I'm stuck as to how to solve it. It doesn't seem to fit anything that I've come across. Thank you fo you help.

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    Start putting $v=h'$ then you can put all the terms depending from $v$ on one side and integrate. – 2017-02-10

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$$\frac{d^2h}{dt^2}=a\left(\frac{dh}{dt}\right)^2+g$$ $$\frac{dh}{dt}=v\quad\to\quad \frac{dv}{dt}=av^2+g$$ Separable ODE : $\quad dt=\frac{dv}{av^2+g}\quad\to\quad t=\int \frac{dv}{av^2+g}$ $$t=\frac{1}{\sqrt{ag}}\tan^{-1}\left(\sqrt{\frac{a}{g}}\:v \right)+\text{constant}$$ $$v=\sqrt{\frac{g}{a}}\tan\left(\sqrt{ag}(t-c_1) \right)$$ $$h=\int vdt=\sqrt{\frac{g}{a}}\int \tan\left(\sqrt{ag}(t-c_1) \right)dt $$ $$h= -\frac{1}{a}\ln\left|\cos\left(\sqrt{ag}(t-c_1) \right) \right|+c_2$$

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    That can also be simplified as $h=- \frac{1}{a}\ln\left|\cos\left(\sqrt{ag}t-c_1\right) \right|+c_2$ – 2017-02-11
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    You are right. Comparing to the last equation in my answer made me find a mistake in my last integral. Now corrected. Tank you for your comment. – 2017-02-11
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    Thank you very much, this makes much more sense. I didn't think of substituting $v=h'$. – 2017-02-12