Explanation

(b) In this process potential energy of the metre stick will be converted into rotational kinetic energy
P.E. of meter stick $= mg\left( \frac{l}{2} \right)$
Because its centre of gravity lies at the middle point of the rod
Rotational kinetic energy $E = \frac{1}{2}I\omega^{2}$
$I =$ M.I. of metre stick about point $A = \frac{ml^{2}}{3}$
$\omega =$ Angular speed of the rod while striking the ground
$v_{B} =$ Velocity of end $B$ of metre stick while striking the ground
By the law of conservation of energy,
$mg\left( \frac{l}{2} \right) = \frac{1}{2}I\omega^{2} = \frac{1}{2}\frac{ml^{2}}{3}\ \left( \frac{v_{B}}{l} \right)^{2}$
By solving we get, $v_{B} = \sqrt{3gl} = \sqrt{3 \times 10 \times 1} = 5.4m/s$