Thread: Physics again

  1. #1
    Kiss the monkey. CodeMonkey's Avatar
    Join Date
    Sep 2001

    Physics again

    I earlier asked a question about physics (as Mr. Squirrel). I was given a brilliant answer in a formula as my answer.


    t=time passed
    i=initial velocity
    h=initial height
    c=current height

    I plugged in a few things, trying to find "t".

    i=0 (falling, so 0)

    And the answer I got was:
    t^2 = -10


    What's wrong with the formula, because I checked the math w/ my teacher, my dad, etc. So it isn't me.

    Any ideas?
    "If you tell the truth, you don't have to remember anything"
    -Mark Twain

  2. #2
    geek SilentStrike's Avatar
    Join Date
    Aug 2001
    heh, first, what the hell kind of notation uses 'i' for velocity? Second, the general formula is like this

    y = y0 + v0t + 1/2 a t^2

    y0 is original position
    y is final position
    v0 is the velocity
    a is the acceleration
    t is the change in time

    A positive v0 implies moving in the positive direction. A positive a implies accelerating in the positive direction. There-in lies your problem, a was negative. Things do not accelerate away from the ground (after you release them, at least).
    Prove you can code in C++ or C# at TopCoder, referrer rrenaud
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  3. #3


    I think it was you that I posted this to. I must have given you the wrong formula:


    where t =time passed
    v=initial velocity
    i=initial height
    y=current height

    I am sorry but I forgot to make 1/2 negative when I have it to u.

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