91Ó°ÊÓ

Diameter of ball (D) = $6.5cm$, terminal speed of ball (v) =$26m/s.$

The formula of terminal speed is given by:

$v_t=\sqrt{\frac{2mg}{\mathrm{ÒÏ}A{C}_D}}.$

where,

m is the mass.

g is the gravitational acceleration.

$\mathrm{ÒÏ}$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 $A=\mathrm{Ï€}{r}^2.$

The radius of the ball is half the diameter:

$$\begin{gathered} r=\frac{d}{2}. \\ =3.13\mathrm{cm}. \\ =0.0313\mathrm{m}. \end{gathered}$$

So the area is,

$$\begin{gathered} A=\mathrm{π}{r}^2. \\ =\mathrm{π}(0.0313\mathrm{m}). \\ =3.1·{10}^{-3}{\mathrm{m}}^2. \end{gathered}$$

And so by using the data of the problem and re-arranging the equation, we can calculate m (the mass) from the terminal velocity:

$$\begin{gathered} m=\frac{v_f^2\mathrm{ÒÏ}A{C}_D}{2g}. \\ =\frac{(18\mathrm{m}/\mathrm{s}{)}^2\left(1.2\mathrm{kg}/{\mathrm{m}}^3\right)\left(3.1·{10}^{-3}{\mathrm{m}}^2\right)(0.47)}{2×9.81\mathrm{m}/{\mathrm{s}}^2}. \\ =0.029\mathrm{kg}. \\ =29g. \end{gathered}$$

The mass of the ball is$29g.$