Answer to Question #258421 in Differential Equations for Max

Question

2. *A body is released from rest and moves under uniform gravity in a medium that exerts a resistance force proportional to the square of its speed and in which the body’s terminal speed is V . Show that the time taken for the body to fall a distance h is

V/g cosh−1 e^(gh/v^2) .

In his famous (but probably apocryphal) experiment, Galileo dropped different objects from the top of the tower of Pisa and timed how long they took to reach the ground. If Galileo had dropped two iron balls, of 5 mm and 5 cm radius respectively, from a height of 25 m, what would the descent times have been? Is it likely that this difference could have been detected? [Use the quadratic law of resistance with C = 0.8. The density of iron is 7500 kgm−3.]

Answer 1

$mv'=mg-cv^2$

terminal velocity:

for v'=0:

$V=\sqrt{mg/c}$

$\frac{dv}{1-\frac{c}{mg}v^2}=gdt$

$\int \frac{dv}{1-v^2/V^2}=g\int dt$

$arctanh\frac{v}{V}=\frac{gt}{V}+C$

$h(t)=\int dv=V\int tanh\frac{gt}{V}dt$

$h(t)=\frac{V^2}{g}ln(cosh(\frac{gt}{V}))$

$t=\frac{V}{g}arccosh(e^{gh/V^2})$

for ball with radius 5 mm:

$V_1=\sqrt{4\pi (0.005)^3\cdot 7500g/(3\cdot 0.8)}=0.22$ m/s

descent time:

$t_1=\frac{0.22}{g}arccosh(e^{25g/0.22^2})=113.65$ s

for ball with radius 5 cm:

$V_1=\sqrt{4\pi (0.05)^3\cdot 7500g/(3\cdot 0.8)}=6.96$ m/s

descent time:

$t_1=\frac{6.96}{g}arccosh(e^{25g/6.96^2})=4.08$ s

It is big difference between descent times, and it can be detected.