A Rather Simple Freefall Question

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SUMMARY

The discussion focuses on calculating the time it takes for a person to fall 4000 feet, given a terminal velocity of 120 mph (176 feet per second) and an acceleration of 32.15 feet per second squared. The initial calculation suggests it takes approximately 27.14 seconds to fall this distance, but after correcting for units, the revised total time is 22.72 seconds. The conversation highlights the importance of considering air resistance, which is not included in the basic calculations but would likely extend the fall time by an estimated 2-3 seconds.

PREREQUISITES
  • Understanding of basic physics concepts, particularly kinematics
  • Familiarity with units of measurement in feet and seconds
  • Knowledge of terminal velocity and its implications
  • Basic understanding of acceleration and its calculation
NEXT STEPS
  • Study the equations of 1-D kinematics for more accurate calculations
  • Learn about air resistance and its effects on falling objects
  • Explore the concept of drag coefficient and its role in terminal velocity
  • Review physics principles related to free fall and acceleration due to gravity
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This discussion is beneficial for students preparing for physics courses, educators teaching kinematics, and anyone interested in the physics of free fall and terminal velocity calculations.

Arepo
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Alright, well I haven't taken physics yet, so I set out to try and figure this problem out on myself. And the problem was, how long does it take for a person to fall 4000 feet if their terminal velocity is 120?

Here is my process of thought to get the answer... Please correct me when I'm wrong (I know I am)

Height = 4000 feet
Increase rate per second = 32.15 feet
Terminal Velocity = 120 mph = 176 feet per second

So to determine how long it takes to get to terminal velocity...

176 / 32.15 = 5.42 seconds to reach TV
(Or is it 176 / 9.8 squared?)

4000 - 176 = 3824 feet left

3824 / 176 = 21.72 seconds to travel that distance at TV

So 21.72 + 5.42 = 27.14 seconds to fall that distance.

I doubt I am right, but it all seems logical and if I messed up, please correct me, thanks a ton!


Edit: Extremely sorry for not posting it in the homework section...
 
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Arepo said:
Alright, well I haven't taken physics yet, so I set out to try and figure this problem out on myself. And the problem was, how long does it take for a person to fall 4000 feet if their terminal velocity is 120?

Here is my process of thought to get the answer... Please correct me when I'm wrong (I know I am)

Height = 4000 feet
Increase rate per second = 32.15 feet
Terminal Velocity = 120 mph = 176 feet per second

So to determine how long it takes to get to terminal velocity...

176 / 32.15 = 5.42 seconds to reach TV
(Or is it 176 / 9.8 squared?)

You're good up to here.

Arepo said:
4000 - 176 = 3824 feet left

Now look at your units here... 176 is ft/s. You can't subtract a velocity from a distance. You need to do 176 * 5.42, and you end up with ft., which is ~954. Then you can do 4000 - 954 = 3046 ft left.

Arepo said:
3824 / 176 = 21.72 seconds to travel that distance at TV

So 21.72 + 5.42 = 27.14 seconds to fall that distance.

I doubt I am right, but it all seems logical and if I messed up, please correct me, thanks a ton!


Edit: Extremely sorry for not posting it in the homework section...

So just correcting that distance, you get...

3046 / 176 = 17.3 seconds falling @ TV

17.3 + 5.42 = 22.72 seconds total


However... this probably wouldn't be very close to the real answer (assuming that the terminal velocity of a human IS 120 ft/s, and they stay perfectly still and rigid) because as you go faster, you experience more air resistance. As you near terminal velocity, your acceleration gradually diminishes. I'm not sure of an exact formula, air resistance isn't until next chapter!

Hope this helps :)
 
You can also look at the section of the thread https://www.physicsforums.com/showthread.php?t=110015 "Basic Equations of 1-D kinematics" for equations useful in solving this problem.

"Increase rate per second = 32.15 feet"

Note: this is acceleration, and it expressed in the units of ft/s/s.
 
Excellent! Thanks man!

But just curious, if you did factor in air resistance roughly what would it be? If you don't feel like doing it, that's fine, the help you provided was adequate.
 
I wish I could tell you, but I haven't learned how to yet. I won't know how to for another week or so. Air resistance starts next chapter. In order to calculate air resistance, we'd need a lot more information (drag coefficient, cross-sectional area, etc). My guess is that it would probably take about 2-3 seconds longer than our theoretical value, but that's just a guess.
 
K, but thanks anyways. I guess I'll just have to wait until I take Physics next year! ^_^
 

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