Solve the equation: $mx''(t)=-mg+c(x'(t))^{2} $
Can anyone solve it? I exchanged $x'(t)$ into $v(t)$
I counted both integrals and I have an answer: $v=\sqrt {\frac {m}{c}}(\frac{2 \sqrt{g}}{1 +/- e^{2 \frac {\sqrt {cg}}{\sqrt m}t+v_0}}-\sqrt{g}) $
Can anyone check if it's correct?
HINT: Write $x' = v$ and $x'' = v'$. The equation is then separable and can be integrated to obtain a relation between $t$ and $v$.