Answer
$|\vec{v} |=27.6$ m/s
Work Step by Step
(b)
The horizontal component of velocity remains unchanged, so at A,
$v_{x}=v_{0x}=v_{0}\cos\theta_{0}=(42.0m/s)\cos 60^{o}=21.3m/s$
The vertcal component is given by Eq.4-23
$v_{y}=v_{0y}\sin\theta_{0}-gt=(42.0m/s)\sin 60^{o}-(9.80m/s^{2})(5.50s)$
$ v_{y}=-17.5 m/s$
$\vec{v} $=$( 21.3m/s)\hat{i}+(-17.5 m/s)\hat{j}$
$|\vec{v} |=\sqrt{( 21.3m/s)^{2}+(-17.5 m/s)^{2}}=27.6$ m/s
