Answer
The maximum height is one-quarter of the range.
Work Step by Step
Let $v_0$ be the initial velocity. We can find an expression for the range:
$x = \frac{v_0^2~sin~2\theta}{g}$
$x = \frac{v_0^2~sin~(2)(45^{\circ})}{g}$
$x = \frac{v_0^2~sin~90^{\circ}}{g}$
$x = \frac{v_0^2}{g}$
We can find an expression for the maximum height $\Delta y$:
$v_y^2 = v_{0y}^2+2a_y\Delta y$
$\Delta y = \frac{v_y^2 - v_{0y}^2}{2a_y}$
$\Delta y = \frac{0 - (v_0~sin~45^{\circ})^2}{(2)(-g)}$
$\Delta y = \frac{v_0^2~(\frac{\sqrt{2}}{2})^2}{2g}$
$\Delta y = \frac{v_0^2}{4g}$
We can see that $\Delta y = \frac{x}{4}$.
That is, the maximum height is one-quarter of the range.
