College Physics (4th Edition)

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.