Terminal Velocity Help: Solving for Upward Force and Fall Time

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SUMMARY

The discussion focuses on calculating the upward force of air resistance and the fall time of a stack of coffee filters dropped from a height. The upward force of air resistance for a single coffee filter is determined to be 0.01764 N, while for a stack of five filters, it is 0.0882 N. The key insight is that the fall time for the stack of filters can be calculated by recognizing that the velocity increases by a factor of the square root of 5, leading to a new fall time of approximately 45 seconds multiplied by 1/√5. This results in a fall time of about 20.12 seconds for the stack.

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  • Understanding of Newton's laws of motion
  • Familiarity with the concept of terminal velocity
  • Knowledge of the relationship between force, mass, and acceleration
  • Basic algebra for manipulating equations
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  • Study the principles of terminal velocity in fluid dynamics
  • Learn about the equations of motion under constant acceleration
  • Explore the relationship between force and velocity in air resistance
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fball558
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terminal velocity help!

Homework Statement


ok here is the problem. i found parts (a) and (b), but don't know part c.
You drop a single coffee filter of mass 1.8 grams from a very tall building, and it takes 45 seconds to reach the ground. In a small fraction of that time the coffee filter reached terminal speed.

(a) What was the upward force of the air resistance while the coffee filter was falling at terminal speed?
Fair = N

(b) Next you drop a stack of 5 of these coffee filters. What was the upward force of the air resistance while this stack of coffee filter was falling at terminal speed?
Fair = N

(c) Again assuming again that the stack reaches terminal speed very quickly, about how long will the stack of coffee filters take to hit the ground? (Hint: Consider the relation between speed and the force of air resistance.)
Fall time is approximately



Homework Equations


a is .01764 N
b is .0882

The Attempt at a Solution


not really sure what to do. at first i thought oh 5 times bigger means 5 times faster
WRONG lol
any help would be great
thanks
 
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(Hint: Consider the relation between speed and the force of air resistance.)

force of air resistance is proportional to velocity^2
 


i know the faster an object moves the more air particles it hits with means more air resistance force. but don't know what to do with this force.
would force = kenetic?
if so i could solve for velocity but i don't know height of the building so don't know how that will help.
 


force of air resistance is proportional to velocity^2

And you know the force increased by 5 times so the velocity would need to increase by...
 


but how is velocity going to give me drop time (in seconds) if i don't know the height of the building. (velocity is in m/s and i don't know (m) of building)
 


But the height of the building does not change. Velocity = distance / time
Since the distance does not change the only thing that can alter velocity is time.
 


ok i still got the wrong answer. can you show me what to do?
 


Fair is proportional to v2
v is proportional to 1/t

Fair changes by a factor of 5.
So velocity will change by a factor of \sqrt{5}
And time will change by a factor of 1/\sqrt{5}

So your answer should be t = 45 x 1/\sqrt{5}
 


thanks i had that my only problem was sig figs.
got to hate them lol
problem has 2 sig. figs. but the answer wanted
4 don't know why ?
 

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