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

S_n=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)