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Terminal velocity is 95%? 
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#1
Jul1814, 06:36 AM

P: 2

Hi forum! Why cant falling objects reach terminal velocity in the real world? When an object drops (W = mg) it will experience drag force upwards. And since drag is proportional to velocity, both force will eventually be equal therefore, the object will not accelerate anymore, hence reaching terminal velocity. So why is it assumed that the object is at terminal velocity when the object has reached 95% of its theoretical terminal velocity? Are they assuming this in an ideal world where there is no air resistance and since in object will reach terminal velocity when t = ∞, and since an exponential function will never reach 0, terminal velocity cannot be reached? Thank you for reading this!! 


#2
Jul1814, 06:40 AM

P: 3,927

If the terminal velocity is just an estimate with +/ 5% accuracy then it makes sense to say that. 


#3
Jul1814, 07:27 AM

P: 907

If you change the drag formula, the decay may no longer be exponential. But (barring fairly exotic drag formulae) the result will still be that actual velocity approaches terminal velocity asymptotically. You'll hit the ground before they can become equal. Still, you can get close enough for practical purposes. 95% is a reasonable figure for "close enough". [If you allow for changes in atmospheric density with altitude, it is possible to have a falling object (momentarily) hit terminal velocity exactly. But that's often a more complex model than the situation warrants  it's easier to pretend that "terminal velocity" is a constant] 


#4
Jul1814, 08:02 AM

P: 2

Terminal velocity is 95%?



#5
Jul1814, 08:28 AM

P: 907

Looking at it from a more simplistic viewpoint, suppose that you jump out of a hot air baloon. You are falling at 0 mph. Your terminal velocity is (let's say) 100 mph. You fall for one second against negligible wind resistance, gaining roughly 20 mph. Now wind resistance is equal to about 20% of gravity. You fall for another second against this wind resistance. This time you gain only 80% of 20 mph. That's another 16 mph for a total of 36 mph. Now wind resistance is equal to 36% of gravity. You fall for another second against this wind resistance. This time you gain only 64% of 20 mph. That's another 12.8 mph for a total of 48.8 mph. Call it 50 mph. Now wind resistance is equal to 50% of gravity. The closer your speed gets to terminal velocity the slower your speed increases. It never quite gets there. The difference decays geometrically. It took you three seconds to go from a 100 mph delta to a 50 mph delta. It will take another three seconds to get to a 25 mph delta. Another three to get to a 12.5 mph delta. Another three to get to a 6.25 mph delta. So after 12 seconds, you're roughly at 95% of terminal velocity. [The difference between the simplistic picture and the differential equation is that the differential equation is the ideal limit as one does the simplistic calculation using smaller and smaller time intervals] 


#6
Jul1814, 01:09 PM

P: 3,927




#7
Jul1914, 03:42 AM

P: 3,109

Related problem...
Stand 20ft from a wall. Every 5 mins walk forwards half the remaining distance to the wall. How long will it take to reach the wall? In theory never. In practice at some point you will decide you are close enough. 


#8
Jul1914, 09:13 AM

Mentor
P: 22,247

Since the equation is exponential and terminal velocity is the asymptotic, it is easy to see mathematically that you never reach it. 95% is just a common, convenient "close enough".



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