If size of two balls are same and they are dropped from same height
Then the net force on two balls are given as
[tex]F_{net} = F_g - F_v[/tex]
here we know that
[tex]F_g [/tex] = weight of the ball
[tex]F_v [/tex] = viscous or drag force of medium
[tex]F_{net} = mg - F_v[/tex]
now by Newton's II law we can find the acceleration of ball
[tex]a = g - \frac{F_v}{m}[/tex]
now if we ignore the medium resistance as it is very small
[tex]a = g[/tex]
so both the balls will have same acceleration
and also by kinematics we can say
[tex]y = \frac{1}{2}gt^2[/tex]
[tex]t = \sqrt{\frac{2y}{g}}[/tex]
so here we can see that since two balls are dropped from same height and the acceleration is also same for both so here it will take same time to reach the ground.