PHYSICS QUESTION!!!!

how do you find the height or time?

is it h=squreroot4.9/t

or h=squareroott/4.9 or is it something else complettly???

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asked by NADINE
  1. You are going to have to describe the situation. Where is the height measured from? Is something falling beginning at time t=0? Does g = 9.8 m/s^2?

    If an object is dropped at time t=0 in a gravitational field where g = 9.8 m/s^2, then the distance it falls in time t is

    h = (g/2)*t^2 = 4.9 t^2

    That looks like "something else comletely".

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    posted by drwls
  2. Well, we seem to be playing "Here is a choice of answers. What is the question?"

    Now if I drop a rock in frictionless air on earth, the acceleration is about 9.8 m/s^2 down
    In that case the speed is about
    v = 9.8 t
    and the distance is about
    h =(1/2)(9.8) t^2
    or
    h= 4.9 t^2

    That means if you know h (the height) and want to know how long it took to fall, t

    t = sqrt (h/4.9)

    From that you can get the speed when it reached the ground at h below the dropping point

    v = 9.8 t
    v = 9.8 sqrt (h/4.9)

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    posted by Damon

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