Answer
2.08 m/s
Work Step by Step
Here we use the principle of conservation of energy to find the speed of point A. (Assume the pivot to have zero gravitational PE)
$mgh=\frac{1}{2}I\omega^{2}+\frac{1}{2}ky^{2}-(1)$
From the drawing, we can get
$y=\sqrt {(0.1\space m)^{2}+(0.2\space m)^{2}}-0.1\space m=0.124\space m$
We know that,
$\omega=\frac{v}{L},\space I=\frac{1}{3}mL^{2},\space h=\frac{L}{2}$
Therefore (1)=>,
$mg\frac{L}{2}=\frac{1}{2}\times\frac{1}{3}mL^{2}\times\frac{v^{2}}{L^{2}}+\frac{1}{2}ky^{2}$
$v=\sqrt {\frac{3(mgL-ky^{2})}{m}}$ ; Let's plug known values into this eqaution.
$v=\sqrt {\frac{3[(0.75\space kg)(9.8\space m/s^{2})(0.2\space m)-(25\space N/m)(0.124\space m)^{2}]}{0.75\space kg}}\approx2.08\space m/s$
