Please turn on javascript in your browser to play chess.
Science Forum

Science Forum

  1. 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?
  2. 02 Jun '13 12:29
    Without air friction and assuming a point mass, about 2.7 meters or roughly 9 feet. With air friction, a little higher depending on what you're wearing, how heavy you are, etc.
  3. Standard member RJHinds
    The Near Genius
    02 Jun '13 15:38 / 2 edits
    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?
    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 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.

    The Instructor
  4. 02 Jun '13 15:54 / 1 edit
    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.

    The Instructor
    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.
  5. Standard member RJHinds
    The Near Genius
    02 Jun '13 16:10
    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.
    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.

    The Instructor
  6. 02 Jun '13 16:53
    Originally posted by RJHinds
    Isn't that what I did?
    Nope.
  7. Standard member RJHinds
    The Near Genius
    02 Jun '13 17:00 / 1 edit
    Originally posted by KazetNagorra
    Nope.
    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
  8. 02 Jun '13 17:29
    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

    The instructor
    Very good. The formula I gave is the fourth one in the list.
  9. Standard member RJHinds
    The Near Genius
    02 Jun '13 17:45 / 1 edit
    Originally posted by KazetNagorra
    Very good. The formula I gave is the fourth one in the list.
    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
  10. 02 Jun '13 18:41
    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

    The Instructor
    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.
  11. Standard member RJHinds
    The Near Genius
    03 Jun '13 02:49 / 2 edits
    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.
    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.

    The Instructor
  12. Standard member sonhouse
    Fast and Curious
    03 Jun '13 13:02 / 2 edits
    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.

    The Instructor
    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.

    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.

    Did you get the part where he said he does physics for a living? This is high school math. I am older than you and remembered that much.
  13. 03 Jun '13 13:16
    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.

    The Instructor
    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?
  14. Standard member sonhouse
    Fast and Curious
    03 Jun '13 13:20
    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?
    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?
  15. 03 Jun '13 13:22
    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?
    See my second post in this thread (mass cancels from both sides of the equation).