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Acceleration again

  1. Sep 7, 2009 #1
    Acceleration again....

    1. The problem statement, all variables and given/known data

    You throw a stone vertically upward, with an initial speed of 6.0 m/s, from a third story office window. The window is 12 m above the ground
    How long is the stone in the air?


    2. Relevant equations

    v=V(not) + at

    x=V(not)t + 1/2at(Squared)

    V(Squared) =V(not) + 2ax

    3. The attempt at a solution
    I listed this

    v = ?
    v(not) = 6.0?
    a = 9.80 (Free fall?!?!?!?)
    t = ?
    x= 12

    I assume I would use the v=V(not) + at to find t, but I do not have v. Which equation would I use to find v?

    Thanks again :smile:
     
  2. jcsd
  3. Sep 7, 2009 #2
    Re: Acceleration again....

    First calculate the time it takes for the rock to travel from the office window upwards to the point where V = 0. You have all the information to calculate this.
     
  4. Sep 7, 2009 #3
    Re: Acceleration again....

    How would I go about doing that?
     
  5. Sep 7, 2009 #4

    kuruman

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    Re: Acceleration again....

    Actually, you can do this in one step. The equation

    y(t) = y0 + v0yt - (1/2)gt2

    gives you the position of the stone above ground at any time t. If that's the case, then let tf be the amount of time the stone is in the air. If you replace any time t with tf in the above equation what should you get for y(tf)? In other words, where is the stone when time equal to the time of flight has elapsed?
     
  6. Sep 7, 2009 #5
    Re: Acceleration again....



    I have no idea what that equation means/is.
     
  7. Sep 7, 2009 #6

    kuruman

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    Re: Acceleration again....

    That is an equation that you wrote down as one of the relevant equations and I also explained in my posting what it means. I used variable y instead of x. Exactly what part of "gives you the position of the stone above ground at any time t" don't you understand?
     
  8. Sep 7, 2009 #7
    Re: Acceleration again....

    Vi = 6.0 m/s
    Vf = 0 m/s
    a = -9.8 m/s^2
    Xi = 0
    Xf = ?
    t = ?

    Use Vf = Vi + at, find out the time it takes for it to get from the window to its maximum height.
     
  9. Sep 7, 2009 #8
    Re: Acceleration again....

    You want to find the time it takes to hit the ground. You don't need to use anything with a Vf, there isn't any reason in this problem for you to need Vf.
    First of all, these two equations are the same x=x(not) + V(not)t + 1/2at(Squared) and y(t) = y0 + v0yt - (1/2)gt2

    That will give you the position of the object time t. You want the positiong to be 0, So substitute and use the quadratic formula.
     
  10. Sep 7, 2009 #9
    Re: Acceleration again....

    It's simpler to do it the other way, it'll be easier for him to understand it.
     
  11. Sep 7, 2009 #10

    rl.bhat

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    Re: Acceleration again....

    2. Relevant equations

    v=V(not) + at

    x=V(not)t + 1/2at(Squared)

    V(Squared) =V(not) + 2ax
    [/QUOTE]
    In the problem the net displacement is in the downward direction, acceleration is in the downward direction and the initial velocity is in the upward direction. In this case take downward direction as positive and upward direction as negative, then the relevant equation becomes
    x = -vo*t + 1/2*g*t2. Substitute the values and solve the quadratic to find t.
     
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