Question

Calculate the speed (in m/s) a spherical rain drop would achieve falling from 4.60 km in the absence of air drag and with air drag. Take the size across of the drop to be 3 mm, the density to be 1.00 ✕ 103 kg/m3, and the surface area to be r2. (Assume the density of air is 1.21 kg/m3.)

A) in the absence of air drag
b) with air drag

Answer 1

The speed of a raindrop falling without air drag can be found using the square root of two times the acceleration due to gravity multiplied by the height. With air drag, the terminal velocity is calculated using the drag force equation involving mass, gravity, drag coefficient, air density, and cross-sectional area.

Step-by-step explanation:

To calculate the speed a spherical raindrop would achieve falling from 4.60 km without air drag, we can use the kinematic equation for free fall under the acceleration due to gravity:

v = √(2gh)

where v is the final velocity, g is the acceleration due to gravity (approximated as 9.81 m/s2), and h is the height (4600 m). Plugging in these values gives us the speed in the absence of air drag.

In the case with air drag, the terminal velocity of the raindrop can be calculated using the formula:

vt = √((2mg)/(CdρA))

where m is the mass of the raindrop, Cd is the drag coefficient (assumed to be 0.47 for a sphere), ρ is the density of air (1.21 kg/m3), and A is the cross-sectional area of the raindrop. To find the mass (m) of the raindrop, we use its density and volume. The cross-sectional area A is πr2, with r being the radius of the spherical raindrop (0.0015 m). Once we have all variables, we can calculate the terminal velocity with air drag.