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Gravity help

  1. Dec 26, 2005 #1
    Didn't know where to exactly post this but I'm wondering how you work out 2 things. No wind resistence etc to be considered please.

    I drop an object, it falls for 1 second, how far will it fall and what speed will it be doing?

    Same for 2 secs and 3 secs and so on? How do I work it out and what's the answers?
     
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  3. Dec 27, 2005 #2

    HallsofIvy

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    If a= -g is constant and the object starts with initial speed 0 (dropped) then v(t)= -at and the distance it falls in t seconds is
    x(t)= (-a/2)t2. Surely if you are asked these questions, you have those formulas in your book?
     
  4. Dec 27, 2005 #3
    I'm a 43 year old arborist who works on trees, I have plenty of books on trees and diseases and can tell you how to look after them, climb them, prune them and remove them.

    We often are faced with knowing a few things in this realm.

    I do not have a physics book and since I left school almost 30 years ago have forgotten any of this stuff. I thought I'd come along here and ask a reasonable question and hopefully get the answer without to much BS.

    You may consider what you have written as being the answer but to me it's a load of wierd symbols and makes no sense, probably similar to a bunch of bontanical tree names to you.

    Please put it in plain english along with the example of the answer so I can use this information in tutorials to other arborists who also ponder the question.

    For example, an 80kg untied climber slips and falls, reaction time is 0.2secs, how far has he fallen by the time he reacts, and if he was 5m up the tree and hits the ground how fast will he be travelling, how long would it take and what would the approx body weight equivalent be at impact.

    These are very real scenarios, besides just getting the answer for a one off example I wanted to know the math behind it so I could figure out multiple scenarios. Our occupation is the most dangerous in the world, people die regularly, and countless are mamed for life.

    Your assistance would be appreciated, afterall this is the PF!
     
  5. Dec 27, 2005 #4

    Tide

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    Ekka,

    If an object starting from rest falls for t seconds then it acquires speed [itex]v = g t[/itex] (the symbol v is used for speed but is derived from the word "velocity"). The symbol "g" is a proportionality constant whose value depends on the system of units you want to use. In English units g = 32.2 ft/s^2 (feet per second squared) so v will be in feet per second and in "standard" units it is 9.8 m/s^2 (meters per second squared) so v will be in units of meters per second.

    Being an arborist, you may also be interested in the speed acquired after falling a given height or distance. In this case, you would use the formula [itex]v = \sqrt {2 g h}[/itex]. If you use English units then you measure the distance fallen, h, in feet while in "standard" units h will be measured in meters.

    The symbol [itex]\sqrt[/itex] is the radical or square root symbol. For example, [itex]\sqrt {36}[/itex] is a number whose square is 36 so [itex]\sqrt {36} = 6[/itex]. You should have a calculator available to evaluate the square root for other numbers.
     
  6. Dec 27, 2005 #5
    First you need to know the units you will be using:
    Since the unit of time you specified is the sec. (second), either
    m/s (meters per second), or f/s (feet per second) will likely be used for speed, and meters or feet will likely be the units you are using for distance. Do you have a preference?

    Then we need to know the rate of acceleration:
    For gravity at the Earth's surface this would be about 32.2 ft/s^2, or 9.81 m/s^2.

    Initial speed at the instant you release the object is zero.

    For a constant rate of acceleration:
    Final speed at the end of 1 sec., is the value given in the rate of acceleration; ~32 f/s, or ~10m/s.
    Distance traveled would result from the average speed over the first sec.; (beginning speed plus ending speed divided by two),
    16 feet or ~5 meters.

    Make a chart, listing for each second:

    The average speed for that second;
    The final speed at the end of that second;
    The distance it fell during that second (as a result of the average speed over that second);
    The total distance, the previous distance plus distance gained.
     
  7. Dec 27, 2005 #6
    Ok

    This is much better. I use metric scale.

    So at the end of 1 second we have travelled 4.9m and have a velocity of 9.8ms ... so what is the formula for this? D is distance in m.

    D=???

    Also,what's the answers for 2,3 and 4 seconds?

    Hallsofivy wrote this x(t)= (-a/2)t2 what does this mean?
     
  8. Dec 27, 2005 #7

    matt grime

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    It is Newton's Laws of Motion that you need to know. If you google for that phrase you will find many explanations, hopefully one of those will have the one you want; we cannot guess what you do know, don't know, or wish to know.

    It would help you to bear in mind, if we're going for information on our situations here, that these are free forums and the people here are volunteers who try their best but by no reasonable expectation should they be able to guess all of the backgrounds and intentions of those asking questions. To most of us HallsOfIvy's post *is* in plain English; would you, if asked about willows on a site called 'botanicals forum' expect to have to explain what salix means, or where slyvania comes from as a suffix? In mechanics it is the defacto standard that initial velocity is u, final velocity v, acceleration a, distance s, and time t. For constant acceleration they are related by several formulae that are undoubtedly recoverable by google, but which are if I remember correctly,

    v=u+at

    v^2=u^2 +2as

    s=ut+a(t^2)/2

    For you u, the initial speed is 0.

    as long as your units are consistent there are no worries, ay 9.8 (metres per second squared) for acceleration.
     
    Last edited: Dec 27, 2005
  9. Dec 27, 2005 #8
    I thought my question was quite clear. I also mentioned I wasn't sure where to post it so if it's in the wrong place, fine, get a moderator to move it.

    Funny, you can come along to our Tree forums, post a pic, get help no worries, last place you'd be sent is Google.

    Fat lot of good this place is then.

    I suppose I will have to go to Google or buy a book, what was I thinking?
     
  10. Dec 27, 2005 #9

    matt grime

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    Your question is very clear, but so is the answer you got. At least in some standard, just not yours.

    I edited the post to include the equations with explanations of the symbols, but the reason I suggested you googling is that that way you can stop when you find something that you know does what you want, rather than having to have people here keep coming up with new explanations of things that are already out there until we hit the one you understand.

    Had you posted your second post first you might have got the explanation you wanted first time.

    Besides, Google is your friend. Here is the second hit for "motion under gravity"

    http://farside.ph.utexas.edu/teaching/301/lectures/node19.html
     
    Last edited: Dec 27, 2005
  11. Dec 27, 2005 #10

    Tide

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    Ekka,

    The correct response is "Thank you for your kind help."

    You should be dealing with people who are offering assistance and lending a hand a little differently than you handle palm trees that you're clearing out. :)
     
  12. Dec 28, 2005 #11
    I used the Distance = Rate * Time equation thus giving me Distance = (9.8m/s^2) * (1 sec) = 9.8 meters

    plug in T = 2,3,4 and so fourth and solve for D given R = 9.8 m/s^2
     
  13. Dec 28, 2005 #12

    arildno

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    Since you don't BOTHER to read the replies (don't you dare say you don't understand them; that's a lie), then good riddance to you.
     
  14. Dec 28, 2005 #13

    Doc Al

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    Just for the record, this is incorrect. "Distance = Rate * Time" would work if the rate were constant, but it isn't. See the other responses for the correct equations. (Furthermore, 9.8m/s^2 is an acceleration, not a speed.)
     
  15. Dec 28, 2005 #14
    Actually the rate of gravity is a constant. Gravity is always 9.8 m/s regardless but I do apologize! I didn't mean to say s^2 just second no squared since squaring the second makes velocity an acceleration, my mistake.
     
  16. Dec 28, 2005 #15

    OK first gravity is a force so by newtons second law assuming no other forces are acting it will cause an acceleration, g is this acceleration and near the surface of earth it is -9.8 m/s^2( meters per seond squared). Second acceleration due to gravity is not a constant is is only about -9.8 near the surface of the earth, if you were much higher this value would be lower due to a greater distance from the center of the earth.
     
  17. Dec 28, 2005 #16

    Doc Al

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    Seems like you're still wrong. The acceleration due to gravity is roughly 9.8 m/s^2, which is a constant (near the surface of the earth). The speed of a falling body is certainly not constant!

    As was explained several times already in this thread, for a body falling from rest:

    The speed (V) in m/s after T seconds is given by:
    [tex]V = 9.8 T[/tex]

    The distance fallen (D) in meters after T seconds is given by:
    [tex]D = 4.9 T^2[/tex]
     
  18. Dec 29, 2005 #17
    Ok, seems to work, thanks for the help, I'll pass on the intelligent comment of arildno. I can read the board without logging in.

    And Tide, point taken but not agreed, some need thicker skin.
     
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