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sunil george Grade: 10
        a body dropped from certain height travels 176.4 m in the last 2 seconds , then find the height from which the body was dropped 
7 years ago

Answers : (2)

askiitiansexpert soumyajit_iitkanpur
8 Points
										

Let T and H be the total time of flight and height respectively.


Then H = 0.5 * g * T^2


Let v m/s be the velocity of the particle at the beginning of last  2 second.


v = g*(T-2)


according  to the problem, 176.4 = v*2 + 0.5 * g  * 2^2


                                           = 4gT -2g


                               => T = (176.4 + 2*9.8)/(4*9.8)


                                       = 5


Hence H = 0.5 * 9.8 * 25 = 122.5

7 years ago
Gireesh Chandra Joshi
18 Points
										

let u be the vel. before the 2 sec. then.........


         s=176.4    a=9.9         t=2s


         s=ut+1/2at2


         176.4=u*2+1/2*9.8*4


         u=78.4 m/s


this u will be the final velocity for the block when dropped from heighest point before last 2 sec....


 then,


       u=0  ,v=78.4   ,a=9.8,  s=?


       2as=v*2-u*2


       on calculating   s=313.6m


       total h=313.6+176.4=490m



7 years ago
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