Question

A spherical raindrop 2.7 mm in diameter falls through a vertical distance of 3950 m. take the cross-sectional area of a raindrop = πr2, drag coefficient = 0.45, density of water to be 1000 kg/m3, and density of air to be 1.2 kg/m3. (a) calculate the speed a spherical raindrop would achieve falling from 3950 m in the absence of air drag.

Answer 1

here we know that volume of the drop is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi (1.35\times 10^{-3})^3$

$V = 1.03 \times 10^{-8} m^3$

now the mass of the drop is given as

$m = \rho V$

$m = 1.03 \times 10^{-5} kg$

now the net buoyancy force on the drop is given as

$F_b = \rho_{air} V g$

$F_b = 1.2 (1.03\times 10^{-8})9.8$

$F_b = 1.21\times 10^{-7} N$

now the net force on the drop is

$F = F_g - F_b$

$F = 1.03\times 10^{-5}(9.8) - 1.21\times 10^{-7}$

now acceleration of the drop is given as

$a = \frac{F}{m}$

$a = 9.79 m/s^2$

now the speed of the drop is given as

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(9.79)(3950)$

$v_f = 278 m/s$