Understanding Free Fall: Explained in Simple Terms | K.I.S.I.S."

AI Thread Summary
Free fall acceleration on Earth is approximately 32 feet per second squared, not 93 feet per second squared, which would lead to incorrect calculations of speed. If Tim were to fall from a mile high, using the correct acceleration, it would take about 18.11 seconds to reach the ground, assuming no air resistance. However, if considering terminal velocity, which is around 93 feet per second, it would take approximately 56.8 seconds to fall a mile. The discussion highlights the importance of understanding the difference between constant velocity and acceleration in free fall scenarios. Clarifying these concepts is essential for accurate calculations in physics.
Grimstone
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I had this elaborate example all set up to lay my question out. and I can hear my x wife in my head. "you over think everything"

So here is my question.
I do not understand free falls 93feet/sec /second.
this sounds to me that Tim would be doing 279 feet per second after only 3 seconds. ??:confused:


If Tim fell 1 mile. how long till he reached ground?

Please. K.I.S.I.S keep it simple I'm stupid.
 
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Grimstone said:
So here is my question.
I do not understand free falls 93feet/sec /second.
The acceleration of a free falling body (ignoring air resistance) is about 32 ft/sec/sec.
 
Here is an online calculator for this stuff, but you have to convert it to SI units, not feet and miles.

http://planetcalc.com/981/
 
I agree with Doc Al's approximate value.

To Grimstone (a small explanation, hopefully I can help in case you are confused about free fall stuff): The meaning of 32 ft/sec/sec is essentially the acceleration of the object in free-fall. This value is simply due to Earth's gravity.

If free-fall acceleration was 93 ft/sec/sec as you said, then yes the object would be doing 279 feet per second after only 3 seconds. (But this value for the acceleration is roughly three times more than what it actually is for Earth).

Lastly, about Tim falling a mile, and how much time this would take: To work this out, you need to do a bit of calculus, or use the equations for constant acceleration (the suvat equations, was the name used when they were taught to me).
 
Perhaps Grimstone meant that terminal velocity, the speed at which air resistance equals the force of gravity so we have a constant speed, is 93 feet per second (not "per second per second). At 93 feet per second, it would take 279/93= 3 seconds to fall 279 feet. At that terminal velocity, it would take Tim 5280/93= 56.8 seconds to fall a mile- almost a minute. Of course, Tim isn't really going to be concerned about that!
 
poor tim. he will always be remembered.
 
BruceW said:
poor tim. he will always be remembered.

as the man who sacrificed himself to create a slightly more interesting math problem.
 
my answer is 18.11 secs??
 
here's my solution
1mi=5280 ft
y=-1/2gt^2
-5280 = -1/2(32.2)t^2

t^2= -5280/(-16.1)

t^2 = 327.95

t=18.11 secs...
 
  • #10
looks good to me.
 
  • #11
killme22 said:
here's my solution
1mi=5280 ft
y=-1/2gt^2
-5280 = -1/2(32.2)t^2

t^2= -5280/(-16.1)

t^2 = 327.95

t=18.11 secs...

You forget, he is at terminal velocity! It is not an acceleration problem, but one of constant velocity.
 
  • #12
Yes, Grimstone's tim is traveling at terminal velocity, and killme22's tim is falling without air resistance. I forgot the earlier posts. Both answers are right really, but it depends on the assumptions made. That's one of the things I like about physics, the assumptions we use can often be explicitly stated. But in philosophy, literature, e.t.c. the assumptions are often vague or not explicitly stated.
 
  • #13
That'll be why Philosophers couldn't put a Rover on Mars - but they could discuss the significance all night.
 
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