# Question #e2c28

Aug 26, 2016

It is assumed that the four balls are released from a height which is sufficient for the balls to attain terminal velocity. Also buoyant force is neglected being small in comparison to the weight of the balls.

Terminal velocity is the constant speed that a freely falling object eventually attains when the resistance of the medium (air in the problem) through which it is falling prevents its further acceleration.
Therefore, net force acting on the falling object is zero.
We have the drag equation for drag force ${F}_{D}$

${F}_{D} = \frac{1}{2} \rho {v}^{2} {C}_{D} A$ ......(1)
where $\rho$ is density of the fluid, $v$ is the relative velocity of the object, $A$ is the area of cross section, and ${C}_{D}$ is the drag coefficient.
For a freely falling body the general equation of force is
${F}_{\text{net}} = m g - \frac{1}{2} \rho {v}^{2} {C}_{D} A$
Using Newtons Second Law we obtain acceleration $a$ as
$a = \frac{m g - \frac{1}{2} \rho {v}^{2} {C}_{D} A}{m}$
$a = g - \frac{1}{2 m} \rho {v}^{2} {C}_{D} A$ ........(2)

At equilibrium when terminal velocity ${v}_{t}$ is attained
${F}_{\text{net}} = m g - {F}_{D} = 0$
$\implies m g = {F}_{D}$
Using (1) we get
$m g = \frac{1}{2} \rho {v}_{t}^{2} {C}_{D} A$
${v}_{t} = \sqrt{\frac{2 m g}{\rho {C}_{D} A}}$ .....(3)
In case of four balls other physical attributes being same and only masses are different we see that
${v}_{t} \propto \sqrt{m}$

As such terminal velocity for a ball with lesser mass will be smaller.
All other things being equal ball with $1 k g$ mass will attain terminal velocity first.