Q. A particle is moving in a smooth curve under gravity and its velocity varies as the actual distance from the highest point. Prove that the curve is a cycloid.
Attempt: The eq. of motion is $$v\frac{dv}{ds}=-g\sin{\psi}, ~~~~~~~~~~\frac{v^2}{\rho}=g\cos{\psi}$$
Given, $v=Cs$,$C=constant.$ I have to show that the path is a cycloid which is of the form $s=4a \sin(\psi)$. First eq implies $\sin{\psi}=-C^2s/g$.
2nd eq gives, $C^2s^2d\psi=g\cos{\psi}~ds$ i.e $-1/C^2\times\frac{1}{s}=1/g ~~ln(sec\psi+tan\psi) +C$ (Since, $\rho=\frac{ds}{d\psi}$)
Please help me to solve the problem.