Chapter 4 - Two-Dimensional Kinematics - Problems and Conceptual Exercises - Page 109: 86

$\frac{H}{R}=\frac{1}{4}tan\theta$

We know that $H=\frac{v_{\circ}sin^2\theta}{2g}$ and $R=\frac{g}{v_{\circ}^2sin2\theta}$ Now $\frac{H}{R}=(\frac{v_{\circ}sin^2\theta}{2g})(\frac{g}{v_{\circ}^2sin2\theta})$ $\frac{H}{R}=\frac{sin^2\theta}{2sin2\theta}$ $\frac{H}{R}=\frac{sin^2\theta}{4sin\theta cos\theta}$ $\frac{H}{R}=\frac{sin\theta}{4cos\theta}$ $\frac{H}{R}=\frac{1}{4}tan\theta$