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Quick Question about Acceleration/Gravity

  1. Oct 5, 2010 #1
    1. The problem statement, all variables and given/known data

    Ok, I have 2 calculations I need help with.

    First: For an object dropped from rest, the distance of fall "d" is related to the time of fall "t" and acceleration due to gravity "a" by

    d = 1/2at2 (2 as in squared)

    So, DISTANCE is 4.42 meters and time of fall for object 1 is .70 seconds and object 2 is .79 seconds

    Lets first get through that one, but just incase, ill add in the 2nd problem I must figure out:

    2)if we measure "d" and "t" we can solve for the accerlation, "a":

    a = 2d/t squared

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 5, 2010 #2
    As you were told recently, you're meant to have an attempt and show people what your stuck on.
    Q1
    2*4.42/(.7^2)=a
    now put that in your calculator as is and it will tell you what the acceleration is!
    If this was part of a lab or something your values are fairly inaccurate..
     
  4. Oct 5, 2010 #3
    Yes it is apart of a lab. Can you help me out? Maybe add something extra to this calculation?

    Yes I thought so too about the variables: should they not be pretty much the same? It was time with a stop watch, so im assuming its human error. Correct?

    Did I do this right for: a = 2d/t squared


    Answer is: a = 4.33
     
    Last edited: Oct 5, 2010
  5. Oct 5, 2010 #4
    nothing extra can be added, certainly this experiment will be very inaccurate but your teacher will expect that. look up human reaction times etc for things to write about.
    Also you could do what I used to do which is to find what the answer should be and mold the results around that, its a bit questionable but it makes your experimental procedure look very accurate.
    eg for this experiment you know that a is 9.81m/s^2 so 9.81=2*4.42/t^2 and from that the time you should have is 0.949 seconds. In this experiment you would NEVER get a result that accurate but perhaps say you timed 0.85seconds or something. Also you should have repeated your drop time at least 5 times for this not twice!
     
  6. Oct 5, 2010 #5

    PhanthomJay

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    Please do not double post.

    https://www.physicsforums.com/showthread.php?t=434973

    Fortunately, Pat666 picked up on this again. Otherwise, if someone else responded, you'd be back to square 1 again.
     
  7. Oct 5, 2010 #6
    Sorry, I did not explain my self properly in my first post, so I thought I would just start a new one. I did not start it because I was not satisfied with the result. Sorry If I broke the rules.

    So....should I continue to add in this thread or go to my old thread?>
     
  8. Oct 5, 2010 #7
    This one
     
  9. Oct 5, 2010 #8
    Thanks,

    Did I do this right for: a = 2d/t squared


    Answer is: a = 4.33
     
  10. Oct 5, 2010 #9
    NO, write out exactly what you put in your calculator
     
  11. Oct 5, 2010 #10
    hmmm..

    2*4.42/(.7^2)=a


    So I went first (.7 ^2) = 0.49

    Then I went 2 x 4.42 = 8.84

    8.84 / 0.49 = 18.04

    Ahh wait, I messed up on first calculation, is the above answer right this time?
     
  12. Oct 5, 2010 #11
    with t=0.7
    a=2*4.42/(.7^2)
    a=18.04m/s^2 correct

    with t=0.79
    a=14.16m/s^2

    Those are terrible values for the acceleration due to gravity on Earth, I would really consider altering your time closer to 0.95s.
     
  13. Oct 5, 2010 #12
    lol, ok so I got that correct.

    Yes, seems my lab partner did not work the stop watch correct.

    We had to drop a wood ball, and a lead ball. Now, it should be the exact same time correct?

    So your saying a better time would be around .95s ?
     
  14. Oct 5, 2010 #13
    only assuming the two balls had close to the same physical dimensions and assuming that you dropped from a low enough height (which you did) that the lighter ball did not reach terminal velocity. Just though I'd mention that because a lot of people seem to think two objects dropped from the same height will always take the same amount of time to fall regardless of mass and shape, this is not true but IN YOUR CASE IT IS.

    Yes 0.95 seconds will give you the exact answer but like I said this experiment will never yield that, I'd go for a time like 0.865 seconds or something but if you do TELL YOUR LAB PARTNER.
     
  15. Oct 5, 2010 #14
    Hey thanks, really appreciate this help!!!

    Just 1 thing I am having alittle trouble doing:

    Question: Use the equation g = G Me/r^2 to calculate the expexted calue of the accelaeration, g (in m/s2). The mass and radius of Earth are 5.7974 x 10^24 kg and 6.37 x 10^6 m, respectively.

    ?

     
  16. Oct 5, 2010 #15
    again its a simple matter of subbing in your constants,
    This formula is derived from Newtons law of gravitation, F=Gm1m2/r^2
    anyway G=6.67*10^-11Nm^2/Kg^2
    so g=6.67*10^-11*5.7974 x 10^24/(6.37 x 10^6)^2
    so g = 9.53m/s^2
    obviously g should be 9.81m/s^2 its just rounding errors.
     
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