Question

84. Calculate the velocity a spherical rain drop would achieve falling from 5.00 km (a) in the absence of air drag (b) with air drag. Take the size across of the drop to be 4 mm, the density to be 1.00 × 103 kg/m3 , and the surface area to be πr2 .

Answer 1

Answer:

a) $v=313.209\ m.s^{-1}$

b) $v_t=3751.79\ m.s^{-1}$

Explanation:

Given:

• height of the raindrop, $h=5000\ m$

a)

$v=\sqrt{2g.h}$

$v=\sqrt{2\times 9.81\times 5000}$

$v=313.209\ m.s^{-1}$

b)

given that:

diameter of the drop, $d=4\ mm=0.004\ m$

density of the air, $\rho=1.18\ kg.m^{-3}$

the terminal velocity is given as:

$v_t=\sqrt{\frac{2m.g}{\rho.A.c_d} }$

where:

m = mass

g = acceleration due to gravity

$\rho=$ density of the medium through which the drop is falling (here air)

A = area normal to the velocity of fall

$c_d=$ coefficient of drag = 0.47 for spherical body

$v_t=\sqrt{\frac{2\times 5\times 9.81}{1.18\times \pi\times 0.002^2\times 0.47} }$

$v_t=3751.79\ m.s^{-1}$