02 Jun '13 12:18

The rate of descent for the T-10D parachute is 22 - 24 feet per second.

What elevation is that equivalent to jumping off of without a parachute?

What elevation is that equivalent to jumping off of without a parachute?

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Fort Gordon02 Jun '13 15:382 edits

I took math and physics about 50 years ago, but if I remember correctly this is an easy problem unless I am over simplifying it. The acceleration of a freely falling body is approximately 32ft per second squared.*Originally posted by USArmyParatrooper***The rate of descent for the T-10D parachute is 22 - 24 feet per second.**

What elevation is that equivalent to jumping off of without a parachute?

So the distance you would fall in one second would be half the 32 ft which is 16 ft. in one second. Since your parachute is slowing you down to 22 to 24 ft per second squared and is maintaining that constant speed, then that would be equivalent to jumping from 11 to 12 feet without the parachute, because all you have to do is divide your speed by 2.

I have never jumped from a parachute, but a 11 or 12 foot jump would not be such an easy fall unless the grass is soft. I recommend getting a better parachute.

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Germany02 Jun '13 15:541 edit

Your computation doesn't make much sense. The correct velocity (assuming no friction, no changes in the location of the center of mass and an initial velocity of zero) can be obtained from conservation of energy; (1/2) * mass * velocity^2 = mass * gravitational acceleration * height.*Originally posted by RJHinds***I took math and physics about 50 years ago, but if I remember correctly this is an easy problem unless I am over simplifying it. The acceleration of a freely falling body is approximately 32ft per second squared.**

So the distance you would fall in one second would be half the 32 ft which is 16 ft. in one second. Since your parachute is slowing you down ...[text shortened]... n easy fall unless the grass is soft. I recommend getting a better parachute.

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Fort Gordon02 Jun '13 16:10

Isn't that what I did? You didn't specify any details. So I might have over simplified it. But that still should be the approximate answer from what information you provided.*Originally posted by KazetNagorra***Your computation doesn't make much sense. The correct velocity (assuming no friction, no changes in the location of the center of mass and an initial velocity of zero) can be obtained from conservation of energy; (1/2) * mass * velocity^2 = mass * gravitational acceleration * height.**

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Germany02 Jun '13 17:29

Very good. The formula I gave is the fourth one in the list.*Originally posted by RJHinds***I didn't realize I was not responding to the first poster. So whatever you say doesn't matter.**

Equations for a falling body.

http://en.wikipedia.org/wiki/Equations_for_a_falling_body

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Fort Gordon02 Jun '13 17:451 edit

We are only concerned about the distance. So the first one is best.*Originally posted by KazetNagorra***Very good. The formula I gave is the fourth one in the list.**

Parachute landing - It is good to know how to fall.

YouTube

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Germany02 Jun '13 18:41

But the time of the fall is unknown a priori. The velocity is given. The time of the fall can be calculated if you know the height (as it happens, it's roughly three quarters of a second), but it's not necessary for this problem.*Originally posted by RJHinds***We are only concerned about the distance. So the first one is best.**

Parachute landing - It is good to know how to fall.

http://www.youtube.com/watch?v=giKo8oKf4GU

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I'm really baffled you would be arguing with someone who does physics for a living about these sort of high school physics problems. But I suppose you're just trolling again. Touché.- Joined
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Fort Gordon03 Jun '13 02:492 edits

The time of the fall does not need to be calculated. It was given as one second. He said the fall rate was 22 - 24 feet per second. Notice that second at the end. I am baffled that you are trying to make a simple problem more complicated than it needs to be. All you need to do is replace the g in the equation with the new fall rate with the parachute on. Simple as that. He did not ask how long it would take for him to fall from that distance under the force of gravity without the parachute.*Originally posted by KazetNagorra***But the time of the fall is unknown a priori. The velocity is given. The time of the fall can be calculated if you know the height (as it happens, it's roughly three quarters of a second), but it's not necessary for this problem.**

I'm really baffled you would be arguing with someone who does physics for a living about these sort of high school physics problems. But I suppose you're just trolling again. Touché.

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slatington, pa, usa03 Jun '13 13:022 edits

I am baffled that you can't read. The question is how far do you fall to have a final velocity of 23 feet per second. You do need to figure the time. And like the man says, if you multiply 32 feet per second times .75 second^2 divide by two you get 9 feet.*Originally posted by RJHinds***The time of the fall does not need to be calculated. It was given as one second. He said the fall rate was 22 - 24 feet per second. Notice that second at the end. I am baffled that you are trying to make a simple problem more complicated than it needs to be. All you need to do is replace the g in the equation with the new fall rate with the parachute ...[text shortened]... m to fall from that distance under the force of gravity without the parachute.**

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So a final velocity of the parachute is the same as falling 9 feet not 11 as you think you calculated. Don't know why you even TRY to fly with the big boys.

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Germany03 Jun '13 13:16

Now you're just taking the piss. The time was not given, the velocity was given. If you're driving 50 miles per hour you're not necessarily driving for an hour. You get that, right?*Originally posted by RJHinds***The time of the fall does not need to be calculated. It was given as one second. He said the fall rate was 22 - 24 feet per second. Notice that second at the end. I am baffled that you are trying to make a simple problem more complicated than it needs to be. All you need to do is replace the g in the equation with the new fall rate with the parachute ...[text shortened]... m to fall from that distance under the force of gravity without the parachute.**

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slatington, pa, usa03 Jun '13 13:20

Hi, I just wondered, I had to sneak up on the answer, just plugging in numbers to the S=AT^2/2 thing. Is there an equation that does it in one swell foop? And the time, T=Sqr root (2S/A) I had to use both equations. Is there a single one that does the job?*Originally posted by KazetNagorra***Now you're just taking the piss. The time was not given, the velocity was given. If you're driving 50 miles per hour you're not necessarily driving for an hour. You get that, right?**- Joined
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Germany03 Jun '13 13:22

See my second post in this thread (mass cancels from both sides of the equation).*Originally posted by sonhouse***Hi, I just wondered, I had to sneak up on the answer, just plugging in numbers to the S=AT^2/2 thing. Is there an equation that does it in one swell foop? And the time, T=Sqr root (2S/A) I had to use both equations. Is there a single one that does the job?**