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A body falls freely from the topof the tower and during the last second of its flight,it falls 16/25th of the whole distance heightof tower is (a) 10.2m. (b) 15.2m. (c)31.25. (d) 33
let total height be h and time be t
therfore in t''th sec, it falls 16h/25
16h/25=5(2t-1) - (hope you know the formula of distance covered in nth second)
solve this for a equation in t and h.
also s=ut +1/2at2
therfore h=5t2
now two equations 2 variables put h=5t2 in the first one. 2 values will come take the positive one and put it in 2nd one.
find the answer.
15.2m
distance travelled in nth second=u+g/2(2n-1)
(c)31.25
Sn=u + {a(2n-1)}/2
16h/25= g(2n-1)/2
h=g(2n-1)*25/32
but, h = ut+1/2gt2
h=1/2g*n2
therefore, 25/32*g(2n-1)=1/2g*n2
50n-25/32=n2/2
50n-25=16n2
16n2 -50n +25=0
16n2 -40n-10n +25=0
8n(2n-5) -5(2n-5)=0
(2n-5)(8n-5)=0
n=5/2 or n 5/8(wrong, because it is less than 1 second)
h=1/2*10*(5/2)2
h=125/4=31.25
(C)
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