A waterdrop of mass 40.1 mg (that is milli- gram) falls in quiet air (no wind).

Question

A waterdrop of mass 40.1 mg (that is milli- gram) falls in quiet air (no wind). Due to the air resistance force, the drop has a maximum velocity called the terminal velocity of magnitude, 14.0 m/s, which is reached high above ground.

a) What would be in N the weight of the rain drop after it reaches terminal velocity and before it reaches ground? Use g = 10 N/kg and take the upward direction as positive. b) What would be in Newtons the weight of the rain drop before it reaches terminal velocity. Use g = 10 N/kg and take the upward direction as positive.

c) What would be in m/s2 the acceleration of the rain drop after it reaches terminal velocity and before it reaches ground? Use g = 10 N/kg and take the upward direction as positive.

d) What would be in m/s the velocity of the rain drop as it hits the ground? Assume that is falls from a height of 10.3 km. Use g = 10 N/kg and take the upward direction as positive.

Explanation / Answer

terminal velocity

v=(2mg/CdA)1/2

where C=0.5 for spherical objects

A is area of cross section

d=density of air

mg is weight of teh drop

part a)

mg=v2CdA/2

part b)

weight of rain drop before it reaches terminal velocity=40.1*10

part c)

after it reaches terminal velocity, taht is constant velocity

its acceleartion will be zero

part d)

it hits the ground with terminal velocity

that is 14 m/s

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