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feather takes T seconds to fall L m from rest (MCQ) Watch

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    If anybody would be kind enough to explain exactly how the answer of this problem is worked out, I would be ever so grateful.
    I understand that this is a suvat question and that
    s = x
    u = 0 m/s
    v = unknown
    a = 9.81 m/s/s
    t = 0.50T
    Now we can use s = ut+0.5at^2 and that'll give me 1.22625T^2. I can't have that as an answer. What must I do now?
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    You haven't used the information that the feather takes a time of T to fall through the distance L. How can you put together an equation that uses this information? And how does it help you once you've done it?
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    (Original post by originaltitle)

    If anybody would be kind enough to explain exactly how the answer of this problem is worked out, I would be ever so grateful.
    I understand that this is a suvat question and that
    s = x
    u = 0 m/s
    v = unknown
    a = 9.81 m/s/s
    t = 0.50T
    Now we can use s = ut+0.5at^2 and that'll give me 1.22625T^2. I can't have that as an answer. What must I do now?
    Do the same thing again, except set s=L and t=T, then solve simultaneously for x.
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    Or just use proportionality

    if s is prop to t squared
    what happens in half the time?
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    Thanks all, I got the correct answer. Thanks very much.
 
 
 
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