@CalvinKhor but see it's easy famous problem that is solvable by smartarse and population was less now it's modern era nearly 8 billions of people thousands have tried
@SMSheikh what is the answer to the rotation question? I think i have solved it
i am getting:
w=$\sqrt(g/l)$
and , the vertical position of end B= $ut - 1/gt^2 - (l/2)sin(wt)$
which is maximum at $t=(\pi + 0.45)\sqrt{l/g}$
My solution:
FIrst part: COM of rod comes back to the ground in time = $2u/g$.Thus, $w(2u/g)=2\pi$. Second part: draw the positon of the rod at some instant of time. Suppose the rod makes an angle $a$ with the horizontal. $a=wt$.If the vertical position of the COM is $y_{c}$ and that of end B is $y_{b}, then $ y_{b}=y_{c}- (l/2)sin(wt)$
divide the entire thing by $\sqrt{gl}$. in the resulting equation, take $sqrt{g/l}$=z. The equation is $\pi-zt=0.5cos(zt)$. let $zt-\pi=x$. then the equation becomes cos(x)=2x
i.e x=0.45. And thus $t=(\pi+(0.45))/z$
and then substitute into the equation of vertical position
i feel this is correct, since it manages to use every single piece of info in the question
