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1-D Kinematics

  1. Dec 1, 2012 #1
    1. The problem statement, all variables and given/known data

    Object dropped from a tower at height of 71.3 metres

    What is the time it takes to hit the ground? What is the velocity when it does?

    gravitational acceleration = 9.8m/s2


    2. Relevant equations

    V = ΔX/Δt

    X = average velocity X time

    3. The attempt at a solution

    I don't know a formula that would enable me to get the time.
     
  2. jcsd
  3. Dec 1, 2012 #2
    Use the WELL-KNOWN equation i.e.

    x = ut + [itex]\frac{1}{2}[/itex]at[itex]^{2}[/itex].
     
  4. Dec 1, 2012 #3
    Try to derive it from formula said by grzz.
    u is initial velocity which is zero as they mentioned it that they dropped.
     
  5. Dec 1, 2012 #4
    Or you can think it in terms of energy mgh=½mv^2 ...etc to solve for the final velocity. On the other hand vy=g*t so you get t.
     
  6. Dec 1, 2012 #5
    I've tried solving for time and this was what I got:

    X= Xo + Vot + 1/2at2

    Vo = 0 (it's at rest)

    Xo = 0 (also at rest)

    a = 9.8m/s2

    X = 71.3m

    71.3 = 0 + 0 + (.5 X 9.8 X t2)

    71.3 = .5 X 9.8 X t2

    71.3 / .5 = 9.8 X t2

    71.3 / .5 / 9.8 = t2

    √(71.3 / .5 / 9.8) = t

    t ≈ 3.81s

    Does this mean that time equals: √X/.5/a ? If initial velocity and initial position equals 0?

    If so, how do you calculate it if the initial velocity and position doesn't equal 0?
     
    Last edited: Dec 1, 2012
  7. Dec 2, 2012 #6

    haruspex

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    Same equation. Plug in the known initial position and velocity and solve the quadratic.
     
  8. Dec 2, 2012 #7
    Not what I meant. I meant how to solve for "t"
     
  9. Dec 2, 2012 #8

    MarneMath

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    Last time I checked, when you drop an object, the initial velocity is 0 and, if you wanted you could make your reference line the point at which you dropped the object and thus have the initial position be 0, but then you have to be careful with your signs.

    Is any of this making sense?
     
  10. Dec 3, 2012 #9

    haruspex

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    That's what I thought you meant, and I thought I answered.
    Xf= Xo + Vot + 1/2at2
    If you know Xf, Xo, Vo and a, plug those in and solve the resulting quadratic to find t. Doesn't matter whether Xo and Vo are zero or nonzero, provided they're known.
     
  11. Dec 3, 2012 #10
    But how can you get Xf if you don't have t? And in the quadratic, I'm not sure which ones would be a, b, or c. I'm thinking in the quadratic formula it's something like.

    t = x

    b = Vi

    a = a

    c = Xf

    so it will then look like: at^2 + Vit - Xf = 0

    I'm trying to figure this out.
     
  12. Dec 3, 2012 #11
    Shouldn't the first term be ½at^2?
     
  13. Dec 3, 2012 #12

    haruspex

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    This whole discussion assumes you know everything in the equation except t.
    In the OP, you had X0 = 0, V0 = 0, Xf = 71.3, a = g, and you solved it happily. To do that, you effectively rearranged the equation to be
    (a/2)t2 + V0t + X0 - Xf = 0
    Then you asked what if X0, V0 are nonzero, to which I answered, just plug those nonzero values into the same equation.
     
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