Answer
$v = 5.42~m/s$
Work Step by Step
Note that the initial height of the center of mass is $0.500~m$
We can use conservation of energy to find the angular speed just before hitting the floor:
$U_0 = K_f$
$mg~h = \frac{1}{2}I~\omega^2$
$mg~h = (\frac{1}{2})(\frac{1}{3}mL^2)~\omega^2$
$\omega^2 = \frac{6~g~h}{L^2}$
$\omega = \frac{\sqrt{6~g~h}}{L}$
$\omega = \frac{\sqrt{(6)~(9.80~m/s^2)~(0.500~m)}}{1.00~m}$
$\omega = 5.42~rad/s$
We can find the speed of the other end of the meter stick:
$v = \omega~r$
$v = (5.42~rad/s)(1.00~m)$
$v = 5.42~m/s$
