Trigonometry (11th Edition) Clone

In part (a), when $h=0$: $D = \frac{v^2~sin~(2~\theta)}{32}$ We can find $D$ when $\theta = 30^{\circ}$ and $v = 36~ft/s$: $D = \frac{(36~ft/s)^2~sin~[(2)(30^{\circ})]}{32}$ $D = \frac{(36~ft/s)^2~sin~(60^{\circ})}{32}$ $D = 35~ft$ The stone will travel 35 feet.