Question

26. Review. (a) Estimate the terminal speed of a wooden sphere (density 0.830 g/cm³ ) falling through the air, taking its radius 8.00 cm and its drag coefficient 0.500 . (b) From what height would a freely falling object reach this speed in the absence of air resistance?

Short answer

The solution is

a) wooden sphere's final velocity as it falls through the air is $\approx 2661.3cm/s$.

b) the absence of air resistance, the free-falling sphere reaches terminal velocity at a height of $36.1m$.

Step-by-step solution

Definition of the terminal velocity

To determine the terminal speed of a wooden sphere falling through air, use the phrase below.

$v_T=\sqrt{\frac{2mg}{D{\rho }_{air}A}}\mathrm{....}\left(1\right)$

Here, m is the sphere's mass, g is gravity's acceleration, A is the wooden sphere's area, D is the drag coefficient, ${\rho }_{air}$is the density of air, and $v_{\gamma }$is the terminal speed.

In terms of density and volume, express the relationship for the sphere's mass as

$m={\rho }_{sphere}V$

Here,${\rho }_{sphere}$ is the sphere's density, and V is the sphere's volume.

In the terminal velocity formula, replace m with sphere${\rho }_{sphere}$ V.

$$\begin{gathered} {\nu }_T=\sqrt{\frac{2{\rho }_{sphere}Vg}{D{\rho }_{air}A}} \\ {V}_T=\sqrt{\left(2{\rho }_{sphere}Vg\right)/\left(D{\rho }_{air}A\right)} \end{gathered}$$

.

Calculating the value of the terminal velocity

(a)

For the terminal speed, rewrite the expression.

$v_T=\sqrt{\frac{2{\rho }_{sphere}Vg}{D{\rho }_{air}A}}$

Substitute$4/3\pi r^{3}forVand4\pi r^{2}for{v}_r.$

$\begin{array}{c}{v}_T=\sqrt{\frac{2{\rho }_{sphere}\left(4\pi r^3/3\right)g}{D{\rho }_{air}}\left(4\pi r^2\right)}\\ =\sqrt{\frac{2r{\rho }_{spher}g}{3D{\rho }_{air}}}\end{array}$

Substitute 8.00 cm for r for 0.830 g/cm³, for ${\rho }_{sphere}$, 980 cm/s² for g, 0.500 for D, 0.001225 g/cm³for${\rho }_{air}$ into the above equation.

$\begin{array}{c}{v}_T=\sqrt{\frac{2\times 8.00cm\times 0.830g/c{m}^3\times 980cm/s^2}{3\times 0.500\times 0.001225g/c{m}^3}}\\ =2661.3279cm/s\\ \approx 2661.3cm/s\end{array}$

As a result, the wooden sphere's final velocity as it falls through the air is .

$\approx 2661.3cm/s$

Calculating the height of the object

(b)

The terminal velocity of the wooden sphere is the final velocity of the wooden sphere when the object freely falls in the absence of air resistance.

In the absence of air, express the relation to find the height at which a freely falling object reaches terminal speed.

$\begin{array}{c}{v}_f^2=v_i^2+2gh\\ 2gh=v_f^2-v_i^2\\ h=\frac{\left(v_f^2-v_i^2\right)}{2g}\end{array}$

Here, h is the object's height, v_f is the sphere's ultimate speed, and v_i is the sphere's initial speed.

Substitute$0m/sfor{v}_{i}and2661.3cm/sfor{v}_{f}and980cm/s^{2}forg$

$\begin{array}{c}h=\frac{{(2661.3cm)}^2-{\left(0\right)}^2}{2\times 980cm/s^2}\\ =3613.529cm\times \left(\frac{1m}{{10}^{2}cm}\right)\\ =36.13529m\\ \approx 36.1m\\ \end{array}$

As a result, in the absence of air resistance, the free-falling sphere reaches terminal velocity at a height of 36.1 m.