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? \( \dot{F}_{g}=0.000651 \mathrm{~N} \times \hat{j} \) b) What would be the weight of the rain drop before it reaches terminal velocity. \( \dot{F}_{g}=0.000651 \) c) What would be the acceleration of the rain drop after it reaches terminal velocity and before it reaches ground? \( \dot{i}=0 \) m/s^2 d) What would be the air resistance force acting on the rain drop after it reaches terminal velocity and before it reaches ground?. \( \dot{F}_{R}=0.000651 \mathrm{~N} \)

Let's analyze the problem step by step based on the given conditions. ### Known Conditions: - Mass of the raindrop: \( m_{1} = 66.4 \, \text{mg} = 66.4 \times 10^{-3} \, \text{g} = 0.0000664 \, \text{kg} \) - Terminal velocity: \( v_{t} = 16.5 \, \text{m/s} \) - Weight of the raindrop: \( \dot{F}_{g} = 0.000651 \, \text{N} \) - Acceleration after reaching terminal velocity: \( \dot{i} = 0 \, \text{m/s}^2 \) - Air resistance force at terminal velocity: \( \dot{F}_{R} = 0.000651 \, \text{N} \) ### a) Weight of the raindrop after it reaches terminal velocity At terminal velocity, the forces acting on the raindrop are balanced. The weight of the raindrop is equal to the air resistance force. Therefore, the weight of the raindrop after it reaches terminal velocity is: \[ \dot{F}_{g} = \dot{F}_{R} = 0.000651 \, \text{N} \] ### b) Weight of the raindrop before it reaches terminal velocity Before reaching terminal velocity, the weight of the raindrop remains the same as it is determined by its mass and gravitational acceleration. The weight is given by: \[ \dot{F}_{g} = m_{1} \cdot g \] Where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \). Let's calculate it: \[ \dot{F}_{g} = 0.0000664 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \] Calculating this gives: \[ \dot{F}_{g} = 0.000651 \, \text{N} \] ### c) Acceleration of the raindrop after it reaches terminal velocity After reaching terminal velocity, the raindrop no longer accelerates. Therefore, the acceleration is: \[ \dot{i} = 0 \, \text{m/s}^2 \] ### d) Air resistance force acting on the raindrop after it reaches terminal velocity At terminal velocity, the air resistance force is equal to the weight of the raindrop. Thus, the air resistance force is: \[ \dot{F}_{R} = 0.000651 \, \text{N} \] ### Summary of Answers: a) Weight after reaching terminal velocity: \( 0.000651 \, \text{N} \) b) Weight before reaching terminal velocity: \( 0.000651 \, \text{N} \) c) Acceleration after reaching terminal velocity: \( 0 \, \text{m/s}^2 \) d) Air resistance force after reaching terminal velocity: \( 0.000651 \, \text{N} \)