Chapter 6 - Section 6.1 - Solving Trigonometric Equations - 6.1 Problem Set - Page 327: 74

$$686.8\ \mathrm{ft}$$

Since $$ h=-16t^2+600 \sqrt{2} t $$ Then $h $ after $ \sqrt{3}$ seconds given by \begin{aligned} h (\sqrt{3})&=-16(\sqrt{3})^2+600 \sqrt{2} (\sqrt{3}) \\ &=-48+600 \sqrt{2} (\sqrt{3}) \\ &= 686.8\ \mathrm{ft} \end{aligned}