A raindrop of mass $$ m_{1}=66.4 \mathrm{mg} $$ (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, $$ v_{t}=16.5 \mathrm{~m} / \mathrm{s} $$, which is reached high above ground. a) What would be the weight of the rain drop after it reaches terminal velocity and before it reaches ground? $$ \dot{F}_{g}=0.000651 \mathrm{~N} imes \hat{j} $$ b) What would be the weight of the rain drop before it reaches terminal velocity. $$ \dot{F}_{g}=0.000651 \quad imes \hat{j} $$ c) What would be the acceleration of the rain drop after it reaches terminal velocity and before it reaches ground? $$ \dot{a}=0 \mathrm{~m} / \mathrm{s}^{\wedge} 2 $$ d) What would be the air resistance force acting on the rain drop after it reaches terminal velocity and before it reaches ground?. $$ \dot{F}_{R}=0.000651 \mathrm{~N} $$

Question

A raindrop of mass $$ m_{1}=66.4 \mathrm{mg} $$ (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, $$ v_{t}=16.5 \mathrm{~m} / \mathrm{s} $$, which is reached high above ground. a) What would be the weight of the rain drop after it reaches terminal velocity and before it reaches ground? $$ \dot{F}_{g}=0.000651 \mathrm{~N} 	imes \hat{j} $$ b) What would be the weight of the rain drop before it reaches terminal velocity. $$ \dot{F}_{g}=0.000651 \quad 	imes \hat{j} $$ c) What would be the acceleration of the rain drop after it reaches terminal velocity and before it reaches ground? $$ \dot{a}=0 \mathrm{~m} / \mathrm{s}^{\wedge} 2 $$ d) What would be the air resistance force acting on the rain drop after it reaches terminal velocity and before it reaches ground?. $$ \dot{F}_{R}=0.000651 \mathrm{~N} $$

UpStudy AI · Accepted Answer

We begin by noting that the mass of the raindrop is

m₁ = 66.4 mg = 66.4 × 10⁻⁶ kg.

Since the gravitational force (weight) is given by

F_g = m₁g

with g ≈ 9.81 m/s², we compute

F_g = 66.4 × 10⁻⁶ kg × 9.81 m/s² ≈ 0.000651 N.

The answers for each part are as follows:

a) After the raindrop reaches terminal velocity (and high above ground) its weight is still given by mg. Thus:

F_g = 0.000651 N (directed downward, i.e., 0.000651 N × ĵ).

b) Before the drop reaches terminal velocity, its weight is also mg. So, irrespective of its speed, the gravitational force remains:

F_g = 0.000651 N (directed downward).

c) At terminal velocity the forces on the raindrop are balanced (air resistance equals the weight), so there is no net force. Therefore, the acceleration is:

a = 0 m/s².

d) Since at terminal velocity the net force is zero, the magnitude of the upward air resistance force must equal the gravitational force in magnitude. Therefore, the air resistance force is:

F_R = 0.000651 N (directed upward).

In summary:

a) Weight after terminal velocity: F_g = 0.000651 N × ĵ  
 b) Weight before terminal velocity: F_g = 0.000651 N × ĵ  
 c) Acceleration at terminal velocity: 0 m/s²  
 d) Air resistance force at terminal velocity: F_R = 0.000651 N (upward)
After reaching terminal velocity, the raindrop's weight is 0.000651 N downward. Before reaching terminal velocity, its weight is also 0.000651 N downward. At terminal velocity, the raindrop's acceleration is 0 m/s². The air resistance force acting on it at terminal velocity is 0.000651 N upward.

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