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A particle is dropped from top of tower. If it falls half of the height of the tower in its last second of journey. Find the time of journey. A particle is dropped from top of tower. If it falls half of the height of the tower in its last second of journey. Find the time of journey.
h = ut +0.5 gt2as u =0 and g =10 m/s2Let total height = HTotal time = TH = 0 5 gT2 = 5 T2Distance travelled in nth second ( last second )Sn = 0.5 g ( n2 — ( n-1)2 ) = 0.5 g ( 2n-1) = H/2So 0.5 gT2 /2 = 0.5 g ( 2n -1) = 5 (2n -1 )H/2 = 5 (2n-1)Put the value of H equal to 5 T2 and n = TT2 /2 = ( 2T-1)T2 = 2 ( 2T-1)T2 = 4T -2T2 -4T +2 = 0Using quadratic solutionsT can be 3.415 Or 0.585Now 0.585 is not possibleSo T = 3.415 second
h = ut +0.5 gt2
as u =0 and g =10 m/s2
Let total height = H
Total time = T
H = 0 5 gT2 = 5 T2
Distance travelled in nth second ( last second )
Sn = 0.5 g ( n2 — ( n-1)2 ) = 0.5 g ( 2n-1) = H/2
So 0.5 gT2 /2 = 0.5 g ( 2n -1) = 5 (2n -1 )
H/2 = 5 (2n-1)
Put the value of H equal to 5 T2 and n = T
T2 /2 = ( 2T-1)
T2 = 2 ( 2T-1)
T2 = 4T -2
T2 -4T +2 = 0
Using quadratic solutions
T can be 3.415 Or 0.585
Now 0.585 is not possible
So T = 3.415 second
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