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

a)
`v=313.209\ m.s^(-1)`

b)
`v_t=3751.79\ m.s^(-1)`

Step-by-step explanation:

Given:

• height of the raindrop,
  `h=5000\ m`

a)

`v=√(2g.h)`

`v=√(2* 9.81* 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{(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{(2* 5* 9.81)/(1.18* \pi* 0.002^2* 0.47) }`

`v_t=3751.79\ m.s^(-1)`