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Homework Help: How is this formula derived?

  1. Oct 29, 2014 #1
    1. The problem statement, all variables and given/known data
    t=2v/g

    the AAMC used this formula to solve a problem that was asking what would happen to time if the gravity becomes g/6. the answer is that the time increases by a factor of 6.

    2. Relevant equations
    v=v0 + at

    3. The attempt at a solution

    I got the wrong answer by thinking that i can use the free fall equation x=1/2 gt2 Thanks in advance.
     

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  3. Oct 29, 2014 #2

    haruspex

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    You need to post the entire original question.
     
  4. Oct 29, 2014 #3
     

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  5. Oct 29, 2014 #4

    haruspex

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    Next time, please try to post the image the right way up.
    But you don't know x, so how can you use that? Did you assume the height reached would be the same in both gravities?
    There are 5 variables in the 5 SUVAT equations. Each equation uses four of them. Given the values of three and the need to determine a fourth, pick the one with those four.
    Which three do you know here, and which is to be determined?
     
  6. Oct 29, 2014 #5
    I'm sorry about that, I kept trying to make it right side up but it wouldn't budge.

    The one I need to determine is the time. The ones I know are acceleration (gravity) and velocity.
    There is the equation v=v0 + 1/2 at2 but I'm not sure how to manipulate it to come out to t=2v/g. When I tried to make it equal to t, I get t=v-v0/a. I don't understand where they get 2v from since the v0 is added on one side, wouldn't it get subtracted if it moves to the other side?
     
  7. Oct 29, 2014 #6

    haruspex

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    There is no such SUVAT equation. Do you mean s= v0 t + (1/2) at2 , or v=v0 + at?
    Also, to what process are you applying this? E.g. the whole process from throwing the ball up to its landing again?
     
  8. Oct 29, 2014 #7
    I'm sorry that was a typo. The equation was v=v0 + at
    I was applying it to the whole process, from when the ball was thrown up to it landing
     
  9. Oct 29, 2014 #8

    haruspex

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    OK, so v0 is the velocity at which it was thrown up. What's the velocity just before it lands?
     
  10. Oct 29, 2014 #9
    the velocity before it lands is v=√2gh
     
  11. Oct 30, 2014 #10

    haruspex

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    OK, you can do it that way if you can figure out h. Alternatively, what do you think the relationship is between the velocity it was thrown up with and the velocity it lands with?
     
  12. Oct 30, 2014 #11
    Would the two velocities by equal? 2v?
     
  13. Oct 30, 2014 #12

    haruspex

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    Not equal, exactly. Remember, a velocity is a vector, it has direction.
    What are you asking?
     
  14. Oct 30, 2014 #13
    I'm trying to figure out how they came up with t=2v/g
     
  15. Oct 30, 2014 #14

    haruspex

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    Yes, I know that, but your last post finished with the question "2v?". I have no idea what you are asking there.
     
  16. Oct 30, 2014 #15
    Oh ok sorry about that, I didn't know if that's where the 2v in the equation came from. If they're talking about the initial v and the v before it lands.
     
  17. Oct 30, 2014 #16

    haruspex

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    OK, let's get back to where we were. What is the relationship between the initial and final velocities? They are not equal because they are not in the same direction, but it is a very simple relationship.
     
  18. Oct 30, 2014 #17
    Are they inversely proportional to each other? Equal but opposite signs?
     
  19. Oct 30, 2014 #18

    haruspex

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    Two very different propositions. Don't guess - which is it? You should be able to pick the right one from everyday experience.
     
  20. Oct 30, 2014 #19
    Ok they are inversely proportional to each other since they are vectors
     
  21. Oct 30, 2014 #20

    haruspex

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    Do you understand what inversely proportional means? It would mean that one of them equals some constant divided by the other. There is no such process as division by a vector.
    If you throw a ball up at 3m/s and catch it at the same height as it comes down, how fast do you think it will be travelling when you catch it?
     
  22. Oct 30, 2014 #21
    It'll be traveling at 3 m/s when I catch it. I'm sorry I guess I meant they are proportional to each other but that would be wrong since they are equal with just opposite directions
     
  23. Oct 30, 2014 #22

    collinsmark

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    Right, they are in opposite directions, so there is a minus sign that needs to fit in there somewhere.

    So plug that into [itex] v_f = v_i + at [/itex] noting that one of the velocities is a negative value (and taking appropriate care in whether the acceleration is up or down [positive or negative], with your chosen conventions).

    There's another way to approach this derivation you don't like working with the velocity formula. Start with

    [tex] y = vt + \frac{1}{2}a t^2 [/tex]

    where [itex] v [/itex] is the initial velocity. You know that the ball finishes at the same location that it starts. In other words, you know that the ball starts at y = 0 and ends at y = 0. Solve for t. (Substitute whatever you need to in for a, keeping attention to whether it is up or down [positive or negative] given your chosen conventions.)
     
    Last edited: Oct 30, 2014
  24. Oct 30, 2014 #23
    Ok so if I let Vi be negative:

    vf = -vi + at

    If I move vi to the other side it becomes 2v

    2v=at
    2v/a=t
    2v/g=t

    is that correct?
     
  25. Oct 30, 2014 #24

    collinsmark

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    That is one valid way to do it, yes. In that convention that you chose, down is positive. Since the ball initially thrown up, you tacked the negative sign on vi (and this assumes that vi itself is a positive number, equal to vf).

    ------
    Edit: Or perhaps even better, you could have kept the equation in it's original form, vf = vi + at, and then made the substitution that vf = -vi. That way the formula remains the same, and it's only the values that change sign.
    -----------

    You also could have done it using the other convention where up is positive and down is negative. In that case a = -g, the initial velocity is positive and the final velocity is negative.

    But if you didn't figure out before-hand that |vi| = |vf| then you could have derived the equation with [itex] y_f = y_i + vt + \frac{1}{2}at^2 [/itex] knowing full well that [itex] y_i = y_f [/itex], using whichever convention that you chose (the convention of whether up or down is positive). So that's yet another way.
     
    Last edited: Oct 30, 2014
  26. Oct 30, 2014 #25
    Thanks so much for your help and for being patient with me. You're awesome!
     
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