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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2007 06:54:59 IST
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The least distance travelled by a upward projected body in any one full second is a)zero b)2.45m c)4.9m d)1.25m explain how
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Ishwarya |
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22 Mar 2007 09:58:50 IST
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u need to know the initilal velocity without which u can not proced
the question is incomplete
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22 Mar 2007 10:10:12 IST
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hey how are u supposed to calculate the ht reached by the the body if initial velocity isn't given??? if u assume the initial vel to be 0, the qn loses its sense...! then obv the distance will be 0 coz the body wouldn't have moved at all..!!! nd are u sure its distance and not displacement? plz reply.....
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***Time has made fools of us yet again...!*** |
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22 Mar 2007 10:14:49 IST
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U know tht vertically projected body and freely dropped body undergo symmetrical motion which r just opposite in nature.....by this, i mean to say that, the pattern of the motion of a vertically projected body is juust the reverse pattern of a freely dropped body.... so, find out min. dist by freely dropped body, which takes place in its first second..... .: min dist = 4.9 m
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22 Mar 2007 12:47:00 IST
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yaar u have taken u = 0 ok but when u will project the velocity in upward direction.u will not be 0 .if it is so then yaar vo haath main he reh jayegi
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22 Mar 2007 13:10:32 IST
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question incomplete
initial velocity not given
at different time intervals it travels different distanses
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ARUN
TRY,TRY,TRY TILL U SUCCEEEEEEEEEEEEEEEEEEED
YOU CAN DO WHATEVER U WANT,JUST TRY
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22 Mar 2007 13:24:05 IST
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Kghedriu is right. Cool job man See I think A.T.Q. what is the min height covered in 1s, so, the minimum height covered will be when the body reaches tthe max. height in 1s. Here t=1....v=o.....a=-g  s=4.9 He only took it otherways
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See there will be two cases. If u>9.8 it will keep on moving upwards in first 1 sec and distance travelled(d) = u-4.9 m which has a minmum of 4.9. But a second case is possible if u (in upward direction) <9.8. In that case it will cover a distance of u^2/(2g) in (u/g) sec and then freely fall for remaining time. Distance travelled in this case will be d=u^2/(2g) +0.5g*(1-u/g)^2 This shows a minima at u=4.9 sec and distance travelled comes out to be 2.45 m. So right answer is 2.45 m.
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Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
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