# 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

Shaurya Gupta
41 Points
11 years ago

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.

Sumedh hedaoo
10 Points
11 years ago

15.2m

GAURAV AGRAWAL
29 Points
11 years ago

distance travelled in nth second=u+g/2(2n-1)

Divyansh Singhvi
20 Points
11 years ago

(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)