Question

(a) Estimate the terminal speed of a wooden sphere (density 0.870 g/cm3) falling through air, if its radius is 8.00 cm and its drag coefficient is 0.500. (The density of air is 1.20 kg/m3.)

Incorrect: Your answer is incorrect.

Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s

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

Incorrect: Your answer is incorrect.

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?Av2, where c is a constant, ? 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?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*(8.0)^3 x 0.87 ; V = 4/3*pi*r^3 , where r = 8 cm, pi = 22/7
m = 1865.85 g = 1.866 kg
and cross sectional area of wooden block A = pi*r^2 = 0.02010 m^2
now from eqn
(1/2)c?Av^2 - mg = 0 terminal vel "v" = ?, A = 0.02010 m^2, c = 0.5, ? = 1.2 kg/m^3
0.5*1.2*0.02010*v^2 = 2*1.866 * 9.81 ; (g = 9.81 m/sec^2)
v^2 = 3035.73
v(terminal) = 55.097 m/sec -------- 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 = ?
(55.097)^2 = 0 + 2*9.81*h
h = 154.726 m ------------------------ Answer