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calculus 2

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An object falls off a 400-foot tower. With g=32ft/sec^2, as at the Earth’s surface, the object would hit the ground after 5 seconds. What would the acceleration due to gravity have to be to make it reach the ground in half the time?

  • calculus 2 -

    should have been g = -32
    then
    v = -32t + c, but when t = 0 , v = 0 , so c = 0

    we have v(t) =-32t
    then
    a(t) = -16t^2 + k, when t = 0 , s = 400 , so k = 400

    so our distance equation is
    s(t) = -16t^2 + 400
    when it hits the ground, s = 0
    0 = -16t^2 + 400
    16t^2= 400
    t^2 = 25
    t = 5
    which was given, and is correct

    if we didn't know the acceleration, let it be a
    our distance equation would have been
    s(t) = (a/2)t^2 + 400
    at half the time, t = 2.5

    0 = (a/2)(6.25) + 400
    a/2 = -400/6.25
    a = -800/6.25 = -128 ft/sec^2

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