Chapter 10 - Rotational Kinematics and Energy - Problems and Conceptual Exercises - Page 327: 74

$v_t=5.3\frac{m}{s}$

As $U_i+K_i=U_f+K_f$ $\implies mgh+0=0+\frac{1}{2}I\omega^2$ $mgh=(\frac{1}{2})(\frac{1}{3}mL^2)\omega^2$ This can be rearranged as: $\omega=\frac{\sqrt{6gh}}{L}$ We plug in the known values to obtain: $\omega=\frac{\sqrt{6(9.81)(0.475)}}{0.95}$ $\omega=5.6\frac{rad}{s}$ Now $v_t=r\omega$ $v_t=(0.95)(5.6)$ $v_t=5.3\frac{m}{s}$