Problem source

A particle is projected from a point $O$ with speed $\sqrt{2gh},$ where $g$ is the acceleration due to gravity. Show that it is impossible, whatever the angle of projection, for the particle to reach a point above the parabola 
\[
x^{2}=4h(h-y),
\]
where $x$ is the horizontal distance from $O$ and $y$ is the vertical distance above $O$. State briefly the simplifying assumptions which this solution requires.

Solution source

The position of the particle is projected at angle $\theta$ is $(x,y) = (v \cos \theta t, v \sin \theta t - \frac12 g t^2)$, ie $t = \frac{x}{v \cos \theta}$,

\begin{align*}
&& y &= x\tan \theta -\frac12 g \frac{x^2}{v^2} \sec^2 \theta \\
&& y &= x \tan \theta -\frac{1}{4h} (1+\tan^2 \theta) x^2 \\
&& 0 &= \frac{1}{4h} x^2\tan^2 \theta - x \tan \theta + \frac{x^2}{4h} +y \\
\Delta \geq 0: && 0 &\leq \Delta = x^2-4\frac{x^2}{4h}\left (\frac{x^2}{4h}+y \right)  \\
&&&=1-\frac{1}{4h^2}(x^2+4hy) \\
\Rightarrow && x^2+4hy &\leq 4h^2 \\
\Rightarrow && x^2 &\leq 4h(h-y)
\end{align*}

We are assuming that there are no forces acting other than gravity (eg air resistance)