Let the body reach the velocity $v$ in time $t$. Since the body is dropped, initial velocity is zero $(u=0).$

\[v=u+gt\] \[\therefore v=gt\]

If the new time is twice the original time, $t’=2t.$

Let the new velocity reached be $v’.$ \[v’=gt’=g(2t)=2(gt)=2v\]

Thus, the new velocity will be twice the original velocity.

[Equations used above are deduced from Motion Under Gravity]