A particle is dropped from top of tower. During its motion it covers `(9)/(25)` part of height of tower in last 1 sec. Then find height of the tower.

Options

A. 100 m
B. `12.5 m`
C. 145 m
D. `167.5 m`

Answer

B

Explanation

Let h be distance covered in t second `rArr h=(1)/(2)g t^(2)`
Distance covered in t th second `=(1)/(2)g(2t-1)`
`rArr (9h)/(25)=(g)/(2)(2t-1)`
From above two equations, `h=122.5 m`