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?
Originally posted by USArmyParatrooperI 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.
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?
Originally posted by RJHindsYour 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.
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.
The Instructor
Originally posted by KazetNagorraIsn'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.
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 RJHindsVery good. The formula I gave is the fourth one in the list.
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
The instructor
Originally posted by KazetNagorraWe are only concerned about the distance. So the first one is best.
Very good. The formula I gave is the fourth one in the list.
Originally posted by RJHindsBut 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.
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
The Instructor
Originally posted by KazetNagorraThe 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.
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.
Originally posted by RJHindsI 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.
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.
The Instructor
Originally posted by RJHindsNow 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?
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.
The Instructor
Originally posted by KazetNagorraHi, 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?
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 sonhouseSee my second post in this thread (mass cancels from both sides of the equation).
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?