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A body dropped from the top of the tower covers a distance of 7x in the last second of its journey where x is the distance covered in first second. How much time does it take to reach the ground?
WE need to use the formulas :-  S = u×t + (1/2)g×t2   and    v = u + gt
where u is initial speed, v is speed after t seconds, g is acceleration due to gravity and S is distance travelled after t seconds

Distance covered in first second = (1/2) g  = x ...............(1)

( In eqn.(1), since the body is dropped from top of tower u = 0 also t = 1 s )

Let the time taken by the body to hit the ground be n seconds

velocity after (n-1) seconds = (n-1)g ...................(2)

distance covered in nth second = (n-1)g+(1/2)g = 7x  ..................(3)

using eqn.(1) and eqn.(3), we have, (n-1)g+(1/2)g = (7/2)g  or  n = 4 seconds
Answered by Thiyagarajan K | 29 Jun, 2019, 08:32: PM
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