## Trigonometry 7th Edition

Magnitude of the horizontal component of the velocity vector in problem 32 is $|\textbf{V}_{x}|=900\,ft/s$. Given that $t= 2\,s$. Horizontal distance=|$\textbf{V}_{x}$|$\times$time$=900\,ft/s\times2\,s=1800\,ft$