A spherical raindrop 1.5 mm in diameter falls through a vertical distance of 500

Question

A spherical raindrop 1.5 mm in diameter falls through a vertical distance of 5000 m. Take the cross-sectional area of a raindrop A spherical raindrop 1.5 mm in diameter falls through a vertical distance of 5000 m. Take the cross-sectional area of a raindrop 2, drag coefficient- 0.45, density of 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 5000 m in the absence of air drag m/s (b) What would its speed be at the end of 5000 m when there is air drag? (Note that the raindrop will reach terminal velocity after falling about 30 m.) m/s

Explanation / Answer

volume of spherical raindrop = 4/3 r^3 = 4/3 * * 0.075 ^3 = 0.00176625 cm3
mass = 0.00176 gram

a ) speed in the absence of air drag.
s = 1/2 g t^2 ---- t^2 = 2 s / g = 5000 * 2 / 9.8 = 1020.4 ---- t = 32 s
v= g * t = 9.8 * 32 = 313.6 m/s

b)
V terminal = 2 m g / C * air * A
A = r^2 =( .75 10^-3) ^2 * = 1.76625 10^-6 m2
m = 0.00176 gram = 0.00176 10^-3 kg
g = 9.8
C = 0.45
air = 1.2 kg /m3

Vt = {(2 * 0.00176 10^-3 * 9.8) / (0.45 * 1.2 * 1.76625 10^-6)} = 36.167
Vt = 6.0139 m/s

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