A body travels half of its total path in the last second of . . .

Question : A body travels half of its total path in the last second of its fall from rest. Calculate the time of its fall.

Doubt by Ishmeet

Solution : 

Let the total time of fall be T. 
Let the total height covered by the body be h. 

Using 
s=ut+½gt²

Case I : When the body cover complete distance
h=(0)T+½gT²
h=½gT² — (1) 

Case II : When the body cover last half distance

Using
Snth=u+½g[2n-1]

h/2=0+½g[2T-1]
h/2=½g[2T-1]
h=g[2T-1] — (2) 

Equating equation (1) and (2) 

½gT²=g[2T-1] 
½T²=2T-1
T²=2[2T-1]
T²=4T-2
T²-4T+2=0

a=1
b=-4
c=2

D=b²-4ac
D=(-4)²-4(1)(2)
D=16-8
D=8

T=[-b±√D]/2a

T=[-(-4)±√8]/2(1)
T=[4±2√2]/2
T=2[2±√2]/2
T=2±√2

T=2-√2
Rejected, time is too small to cover the total distance. 

T=(2+√2) sec
T=2+1.414
T=3.414 sec

Hence, the total time of fall would be 3.414 sec.