Answer:
![v = √(3gL)](https://img.qammunity.org/2020/formulas/physics/college/ofk26duzeno8c80q80dbhdqywb2lnj9hsv.png)
Step-by-step explanation:
As we know that length of the rod is L and mass is M
so here it one end of the rod is stationary and other end of the rod is rotating about one end
then we will have energy conservation to find the total rotational kinetic energy of rod about its one end is given as
![mg(L)/(2) = (1)/(2)I\omega^2](https://img.qammunity.org/2020/formulas/physics/college/y03zi7czt7hr1i14upnxr1sxm4t8e74e7w.png)
![mg(L)/(2) = (1)/(2)((mL^2)/(3))\omega^2](https://img.qammunity.org/2020/formulas/physics/college/dy8rp322hh70mpbrsk8t4iyg6ddao446mn.png)
![g = (L)/(3) \omega^2](https://img.qammunity.org/2020/formulas/physics/college/5ucd6t8bvnpdguw9zz6fioqaqqdvqxs2jz.png)
![\omega = \sqrt{(3g)/(L)}](https://img.qammunity.org/2020/formulas/physics/college/hrovgbo0o65xjdzjjhcunaibqi8wfmc0kd.png)
so the linear speed of the other end of the rod just before it hit the ground is given as
![v = L\omega](https://img.qammunity.org/2020/formulas/physics/college/po4kjory0rq4vz2o4c1q2u4ltdlji3co3w.png)
![v = √(3gL)](https://img.qammunity.org/2020/formulas/physics/college/ofk26duzeno8c80q80dbhdqywb2lnj9hsv.png)