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![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Jun 2008 23:21:43 IST
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Two bodies r projected simultaneously wid sam e velocity of 19.6m/s from top of a tower one vertically down nd other in vertically upwards.what is d time gap as they reach d ground?
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2 Jun 2008 23:28:52 IST
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itz 4 sec...try 2 give me explaination...
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2 Jun 2008 23:33:42 IST
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Ya..It is 4s.Simple,the body thrown upward again reaches that point with same velocity and behaves as the particle thrown downward.So,The time difference is time of flight when thrown vertically up till the same point which is 2u/g=4s.
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MAKING A MISTAKE IS HUMAN BUT REPEATING IT IS IDIOTIC. |
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2 Jun 2008 23:34:53 IST
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ur absolutely ri8 sandeep
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2 Jun 2008 23:36:17 IST
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see ravi
its easy to see that when body projected up comes back to projected position its speed is 19.6m/s downwards.
now from this instant it will follow other body's trajectory in same way as 2nd body do.
hence the only time gap is due to above mentioned path, that is clearly 4 sec.
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