Question

Calculate the speed (in m/s) a spherical rain drop would achieve falling from 1.20 km in the absence of air drag and with air drag. Take the size across of the drop to be 2 mm, the density to be 1.00 ✕ 103 kg/m3, and the surface area to be πr2. (Assume the density of air is 1.21 kg/m3.)

Answer 1

(a) In the absence of air drag

$v=\sqrt{2gh}\\where\ g=9.8\ m\ sec^{-1}\\h=1200\ m\\v=\sqrt{2\times9.8\times1200}=153.36\ m\ sec^{-1}$

(b) In the presence of air drag

net force at any time is given by

$ma=mg-\frac{1}{2}\rho _{air}C_{sphere}v^2A_{drop}$

$m=V\times\rho_{r}=\frac{4\pi}{3}\times r^3\times\rho{r}=0.004188 kg \\A_{drop}=\pi r^2=0.0000031416\ m^2\\C=0.45$

$a=9.8-{\frac{0.5}{0.004188}\times1.21\times 0.45\times 0.0000031416\times v^2}$

$a=9.8-0.00075v^2$

$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}\times v$

$\frac{dv}{dx}=\frac{9.8-0.00075v^2}{v}=\frac{9.8}{v}-0.00075v$

$\overset{v}{\int}\frac{1}{(\frac{9.8}{v}-0.00075v)}dv=\overset{1.2}{\int} dx$

$-\frac{2000}{3}\times ln(|3v^2-39200|)=1.2$

$ln(|3v^2-39200|)=-0.0018$

$3v^2-39200=1\\3v^2=39201\\v=114.31\ m\ sec^{-1}$