Question

A 6.5-cm-diameter ball has a terminal speed of 18 m/s. what is the ball's mass? use 1.2 kg/m3 for the density of air at room temperature.

Answer 1

The terminal velocity is the highest velocity that an object  can reach when falling through a medium opposing a drag force to the motion, and it is given by
$v_t = \sqrt{ \frac{2 mg}{\rho A C_D} }$
where 
m is the mass
g is the gravitational acceleration
$\rho$ is the density of the medium
$A$ is the projected area of the object
$C_D$ is the drag coefficient

We can consider the ball as a perfect sphere, so its drag coefficient is $C_D=0.47$. The projected area is the cross-sectional area of the ball, so
$A=\pi r^2$
the radius of the ball is half the diameter:
$r= \frac{d}{2}=3.13 cm=0.0313 m$
so the area is
$A=\pi r^2 = \pi (0.0313 m)=3.1 \cdot 10^{-3} m^2$

And so by using the data of the problem and re-arranging the equation, we can calculate m (the mass) from the terminal velocity:
$m= \frac{v_t^2 \rho A C_D}{2 g}= \frac{(18 m/s)^2(1.2 kg/m^3)(3.1 \cdot 10^{-3} m^2)(0.47)}{2 \cdot 9.81 m/s^2}=0.029 kg$