Question

A 570-g squirrel with a surface area of 880 cm2 falls from a 5.2-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the cross-sectional area of the squirrel can be approximated as a rectangle of width 11.2 cm and length 22.4 cm. Note, the squirrel may not reach terminal velocity by the time it hits the ground. Give the squirrel's terminal velocity, not it's velocity as it hits the ground.)

Answer 1

Answer : v = 14.28 m/s

Explanation :

It is given that,

mass of squirrel, $m=570\ g$

Surface area, $A=880\ cm^2=88\times 10^{-3}$

Height, $h=5.2\ m$

Terminal velocity is given by :

$v_t=\sqrt{\dfrac{2mg}{\rho AC}$

where, $\rho$ is density of fluid which is falling and it is given by

$\rho=\dfrac{m}{V}$

since, $volume=area\times height$

so, $\rho=\dfrac{0.57\ Kg}{0.088\ m^2\times 5.2\ m}$

$\rho=1.24\ Kg/m^3$

A is the surface area of squirrel.

C is the drag coefficient.

The surface area facing the fluid is given by : $A_f=\dfrac{0.088\ m^2}{2}=0.044\ m^2$

so, terminal velocity is :

$v_t=\sqrt{\dfrac{2\times 0.57\ Kg\times 9.8\ m/s^2}{1.24\ Kg/m^3\times 0.044\ m^2\times 1}}$

$v_t=\sqrt{ 204}$

$v_t=14.28\ m/s$

So, the terminal velocity of squirrel is 14.28 m/s.