Question

Estimate the terminal speed of a wooden sphere (density 0.850 g/cm3) falling through air, if its radius is 8.50 cm and its drag coefficient is 0.500. (The density of air is 1.20 kg/m3.)

Answer 1

Fair + F(downwards) = 0 , i.e., --------------------------1)
1/2*c?Av2 - mg = 0 gravitational force acting on wooden sphere will be counter acted by Fair to to reach terminal velocity using eqn 1)
Now as we know (mass of wooden block = Volume of sphere x Density)
m = (4/3)*pi*(8.5)^3*0.85 ; V = 4/3*pi*r^3 , where r = 8.5 cm, pi = 22/7
m =2186.57 g = 2.1865 kg
and cross sectional area of wooden block A = pi*r^2 = 0.0227 m^2
now from eqn
(1/2)c?Av2 - mg = 0 terminal vel "v" = ?, A = 0.0227 m^2, c = 0.5, ? = 1.2 kg/m^3
0.5*1.2*0.0227*v^2 = 2*2.1865 * 9.81 ; (g = 9.81 m/sec^2)
v = sqrt((2*2.1865 * 9.81)/(0.5*1.2*0.0227))
v(terminal) = 56.12 m/s

terminal speed of a wooden sphere = 56.12 m/s