A body dropped from top of tower falls through 60 m during the last 2 seconds of its fall. The height of tower is `(g = 10m//s^(2))`

Options

A. 95 m
B. 80 m
C. 90 m
D. 60 m

Answer

B

Explanation

U = O, velocity attained by ball at t = 2 second
`v=u+a t=0+g(t-2)`
distance traveled by ball in last 2s is given by
`h_(1)=v t+(1)/(2) at^(2)`
`60=g(t-2)xx2+(1)/(2) g(4)=20(t-2)+20`
at 4S, Height of tower
`=(1)/(2)g t^(2)=(1)/(2)xx10xx 4^(2)=80m`