# Show that the profile of the hill is a cycloid

I'm struggling with this problem:

From George Simmons_ Differential Equations

At sunset a man is standing at the base of a dome-shaped hill where if faces the setting sun. He throws a rock straight up in such a manner that the highest point it reaches is level with the top of the hill. As the rock rises, its shadow moves up the surface of the hill at a constant speed. Show that the profile of the hill is a cycloid.

I've tried different things and I showed that the $x$ component and the $y$ component of the shadow should move with constant speed, but nothing really other than that.

• Cross-posted on Phys.SE: physics.stackexchange.com/questions/133378/… – zeta Aug 31 '14 at 21:57
• I didn't really know where to post it. Because it's a physics question on a Differential Equations Book – Keith Aug 31 '14 at 22:02

Let $u$ be the constant speed of the shadow on the hill. Choose the origin of time to be such that the vertical velocity (virtual before the time of projection) of the projected particle is $u-gt$ at time $t$ (and so equal to $u$ at $t=0$). At time $t$, let $\psi$ be the angle between the horizontal and the tangent to the hill at the shadow point, and let $s$ be the distance along the profile of the hill to the point from some origin. Then$$\frac{\mathrm ds}{\mathrm dt}=u\quad\text{and}\quad u\sin\psi=u-gt.$$By integration,$$s=\frac{u^2}g(1-\sin\psi),$$where we have chosen the origin of $s$ to be such that $s=0$ when $\psi=\frac12\pi$ and $t=0$. The above equation is the $s$–$\psi$ equation of a cycloid*, whose (local) summit is given by $(s,\psi)=(u^2/g,0)$, corresponding to $t=u/g$. Let us measure any height $y$ vertically from where $s=0$ on the cycloid. Then the hill is of height $u^2/g$. The projectile just attains this height if its general height equation is$$y=\frac{u^2}{2g}+ut-\frac{gt^2}2,$$where the constant term is determined by the given condition $y=u^2/g$ when the vertical velocity is zero, namely when $t=u/g$.
*The parametric cartesian equivalent form of the cycloid equation is$$x=a(\theta-\sin\theta),\qquad y=a(1-\cos\theta),$$where $a=\frac12u^2/g,\,$ $\theta=\pi-2\psi$, and the origin for $x$ and $y$ is the same as the origin for $s$.