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

1. 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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2. 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?
3. (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.
4. Or just use proportionality

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

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