If v is the velocity of the rod at any instant, the emf induced in it will be Bvl. And as the resistance of the rod is R, the induced current will be

$I=\frac{e}{R}=\frac{Bvl}{R}$

Due to this induced current a force will act on the rod which will oppose its motion.

$F_M=iBl=\frac{B^2{l}^{2}v}{R} ....\left(i\right)$

So if T is the tension in the string, equation of motion of rod and mass m will be

$T-F_M=0\times a\text{ and }mg-T=ma$

So eliminating T between these

$mg-F_M=ma\text{ i.e., }a=g-\frac{F_M}{m}$

and on substituting the value of F_M from Eq. (i)

$\begin{array}{l}a=g-\frac{B^2{l}^2}{mR}v\\ \text{ i.e., }a=g-\frac{v}{t}\text{ with }\tau =\frac{mR}{(Bl{)}^2} .......\left(ii\right)\end{array}$

So the rod will achieve terminal velocity v_T when a= 0,

$\text{ i.e., }g-\frac{v_T}{\tau }=0$

$\text{ i.e., }{v}_T=\tau g=\frac{mgR}{(Bl{)}^2} .......\left(iii\right)$

$\text{ and so for }v=\frac{1}{2}{v}_T=\frac{1}{2}\tau g,\text{ from Eq. (ii), }$

$a=g-\frac{1}{\tau }\left(\frac{1}{2}\tau g\right)=\frac{1}{2}g=5\mathrm{m}/{\mathrm{s}}^2$