Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.2 - Integrals of Vector Functions; Projectile Motion - Exercises 13.2 - Page 753: 22

$$ 55.4 \mathrm{ft}$$

Since $$y=y_{0}+\left(v_{0} \sin \alpha\right) t-\frac{1}{2} g t^{2}$$ Then $$ y=32+16 t-16 t^{2}$$ The ball hits the ground when $y=0 $ $$\Rightarrow 0=32+16 t-16 t^{2} $$ Hence $ t=-1$ or $t=2 t=2 \sec$ since $t>0 ;$ thus,\begin{align*} x&=\left(v_{0} \cos \alpha\right) t\\ x&=(32 \mathrm{ft} / \mathrm{sec})\left(\cos 30^{\circ}\right) t\\ &=32\left(\frac{\sqrt{3}}{2}\right)(2) \approx 55.4 \mathrm{ft} \end{align*}