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Math Help - initial conditions the hammer and the feather

  1. #1
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    initial conditions the hammer and the feather

    A hammer and a feather are dropped at the same time from 4 feet, in a under a vacuum by an astronaut under the moons gravity.....

    d2s/dt2=-5.2ft/sec^2

    initial conditions: ds/dt=0 and s=4 when t=0.

    So, I find ds/dt, is -5.2T+c, and v(0)=4.

    So v(t)=-5.2t+c=0

    now s(t)=(-5.2/2)*t^2+cx+x

    I want to find t when S(t)=0

    4=(-5.2/2)0^2+kt+c
    c=4

    I want t when s(t)=0.

    0=(-5.2/2)t^2+kt+4

    what now?
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  2. #2
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    Re: initial conditions the hammer and the feather

    Quote Originally Posted by StudentMCCS View Post
    A hammer and a feather are dropped at the same time from 4 feet, in a under a vacuum by an astronaut under the moons gravity.....

    d2s/dt2=-5.2ft/sec^2

    initial conditions: ds/dt=0 and s=4 when t=0.

    So, I find ds/dt, is -5.2T+c, and v(0)=4. no, v(0) = ds/dt when t = 0

    So v(t)=-5.2t+c=0 v(t) = -5.2t

    now s(t)=(-5.2/2)*t^2+cx+x s(t) = -2.6t^2 + C

    s(0) = 4

    4 = -2.6(0)^2 + C

    s(t) = -2.6t^2 + 4
    corrections made
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