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Acceleration due to gravity !

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data
    I was given a lab using the logger lite and go motion probe. After collecting my data, wich is only velocity, time and position, i was asked to calculate the acceleration due to gravity. We have never gone over this and the only information i have is to due with what i said before.
    We do know that acceleration is -9.81 though

    2. Relevant equations
    Is there anyway to find out the acceleration due to gravity for this ! I don't understand how i am suppose to find this out !


    3. The attempt at a solution
     
  2. jcsd
  3. Oct 3, 2011 #2
    If acceleration is constant, then v = at. Try graphing the velocity versus time and looking for the slope.
     
  4. Oct 3, 2011 #3
    For a lab, remember that you don't "know" gravity is [itex]-9.81 \frac{m}{s^2}[/itex] since that is what you are told to find.

    In the most basic sense, what is acceleration? Try listing a few different ways to describe acceleration (i.e. different equations).

    After making the list (make sure you show units in your descriptions), can you use anything from your data to construct this?
     
  5. Oct 3, 2011 #4
    daveb ! thank you for replying so fast ! but my v-t graph has multiple places to find the sole. it starts at 0, accelerates up, then de-celerates ! where do i find my slope?
     
  6. Oct 3, 2011 #5
    that is what i ended up doing, yet i found an exceleration of -1.04m/s2 ! urg that means the percent error is 89! does that even make sens if the ball i threw in the air was a volly ball ?
     
  7. Oct 3, 2011 #6
    daveb; v(t)=at+v0. The constant is actually important, if this isn't a simple freefall-experiment that is.

    integrating this equation with time yields, s(t) = 0,5at^2 + v0*t.
    if you know the initial velocity, the time the journey took and how long it travelled, can't you solve for a?
     
  8. Oct 3, 2011 #7
    Think about where the ball was at each point in time. After it leaves your hands, what forces are acting on it? I don't think that after it left your hand it was accelerating up, but rather just had a positive velocity at that time.
     
  9. Oct 3, 2011 #8
    this is what i have been looking for!! but my teacher hasn't explained what s(t) = 0,5at^2 + v0*t means -_- would you mind explaining this to me?
     
  10. Oct 3, 2011 #9
    it as an equation that describes how long your object travels, s, when you plug in acceleration and time. and initial velocity, v0. :)

    Or did you ask how I got there?
     
  11. Oct 3, 2011 #10
    Thank you :) but now when rearanging my eqation, i have 2 t, do i replace both with the same number ?!
     
  12. Oct 3, 2011 #11
    first of all, what is your initial velocity? if you simply "dropped" the ball, it is 0 and you lose a term.
     
  13. Oct 3, 2011 #12
    no, the ball was thrown then catched, but i see what you're saying , so my v1 is 0, v2 0.157 m and time is 1.8s so the acceleration is 1.045 m/s^2 ? ahh so confusing!
     
  14. Oct 3, 2011 #13
    no matter. you get one term of t^2 and one term of t, but this pose no problem. As far as I've understood you are solving for a, and the value of t is known?

    if it doesn't work, give me all your experimental data and explain what you did in the lab.
     
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