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![[Post New]](/templates/default/images/icon_minipost_new.gif) 12 Jul 2007 20:44:37 IST
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a body is thrown up with a finite speed is caught back after 4 sec. find the speed of the body with which it is thrown up.(g=10m\s2)
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is the ans 20 m/s
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woods are lovely, dark and deep.....
but i have promises to keep....
and miles to go before i sleep....
and miles to go before i sleep.... |
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12 Jul 2007 21:09:00 IST
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diplacement = 0
s=ut + 1/2 at^2
0 = 4u - 1/2 * 10*16
solving we get u= 20 m/s
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woods are lovely, dark and deep.....
but i have promises to keep....
and miles to go before i sleep....
and miles to go before i sleep.... |
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13 Jul 2007 20:12:16 IST
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another method:
time of ascend =time of descend = 4/2 = 2secs
distance covered during ascent=distance covered during descent=s
distance covered during ascent s = ut + 1/2 at2
= 2u+ 1/2 x -10 x 4
= 2u - 20 -----------1
distance covered during descent s = 1/2 at2
= 1/2 x 10x 4
= 20 -------------2
from 1 and 2
2u -20 = 20
2u = 40
u = 20 m/s
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13 Jul 2007 21:14:55 IST
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As the body is caught back after 4sec the time of flight = 4sec
2 u/g=4sec
taking g=10 2u/10=4
2u=40
u=20m/sec
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13 Jul 2007 23:21:37 IST
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Excellent work everyone
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13 Jul 2007 23:27:49 IST
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clever man ..... elessar....
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salman khan |
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14 Jul 2007 13:58:26 IST
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let it go i got the answer ok
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21 Jul 2007 22:24:22 IST
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time of flight, t = 2 u / g
u = g t / 2
= 20 m/s
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21 Jul 2007 23:41:44 IST
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actually...as it is already caught back .....time taken by it to travel in air is nothing but time of flight which is t=2u/g and from data g=10,t=4 => 4=2u/10 which gives us that u=20m/sec.... cheers.!!!
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dont try to be over smart.... |
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