The correct option is D 125 m
Let the height of tower be h and the body takes time t to reach the ground when it falls freely.
Using second equation of motion, we have
h=ut+12gt2 (Take downward direction as positive)
⇒h=12gt2 (∵ u=0 during free fall) .…(i)
In the last second, i.e., in tth second the body travels 0.36h. Thus, in |t−1| seconds, it travels h–0.36h=0.64h
So, we get, 0.64h=12g(t−1)2 ....(ii)
On solving equ. (i) and (ii), we get
⇒h0.64h=12gt212g(t−1)2⟹t=5 s
Substituting in equation (i), h=125 m