Question

(a) Estimate the terminal speed of a wooden sphere (density 0.790 g/cm3) falling through air, if its radius is 7.50 cm and its drag coefficient is 0.500. (The density of air is 1.20 kg/m3.)

[ ] m/s

(b) From what height would a freely falling object reach this speed in the absence of air resistance?

Answer 1

Experiments show that the force of air friction on a falling object such as a skydiver or a feather can be approximated fairly well with the equation |Fair | = c*rho*Av2, where c is a constant, rho is the density of the air, A is the cross-sectional area of the object as seen from below, and v is the object's velocity. Predict the object's terminal velocity, i.e., the final velocity it reaches after a long time.

Fair + F(downwards) = 0 , i.e., --------------------------1)

1/2*c*rho*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*(7.5)^3 x 0.79 ; V = 4/3*pi*r^3 , where r = 7.5 cm, pi = 22/7

m = 1396 g = 1.396 kg

and cross sectional area of wooden block A = pi*r^2 = 0.01767 m^2

now from eqn

(1/2)c*rho*Av2 - mg = 0 terminal vel "v" = ?, A = 0.01767 m^2, c = 0.5, rho = 1.2 kg/m^3

0.5*1.2*0.01767*v^2 = 2*1.396 * 9.81 ; (g = 9.81 m/sec^2)

v(terminal) = 50.83m/s (Answer )

b) Suppose there is no resistance and object is falling strictly under gravitational force

we have v^2 = u^2 + 2gh (3rd law of motion, where v = final vel = v(terminal) , u = 0, h = ?

(50.83)^2 = 0 + 2*9.81*h

h = 131.66 m ------------------------ Answer