## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 4 - Problems - Page 153: 81

#### Answer

Just before impact, the angle $\theta$ below the horizontal of the velocity vector is $63.4^{\circ}$

#### Work Step by Step

We can find the time of flight: $h = \frac{1}{2}gt^2$ $t = \sqrt{\frac{2h}{g}}$ We can find an expression for $v_y$ just before impact: $v_y = v_{0y}+at$ $v_y = 0+gt$ $v_y = g~(\sqrt{\frac{2h}{g}})$ $v_y = \sqrt{2gh}$ We can find an expression for $v_x$: $h = v_x~t$ $v_x = \frac{h}{t}$ $v_x = \frac{h}{\sqrt{\frac{2h}{g}}}$ $v_x = \frac{\sqrt{2gh}}{2}$ We can find the angle $\theta$ below the horizontal of the velocity vector just before impact: $tan~\theta = \frac{v_y}{v_x}$ $tan~\theta = \frac{\sqrt{2gh}}{(\frac{\sqrt{2gh}}{2})}$ $tan~\theta = 2$ $\theta = tan^{-1}~(2)$ $\theta = 63.4^{\circ}$ Just before impact, the angle $\theta$ below the horizontal of the velocity vector is $63.4^{\circ}$

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