A particle is dropped the top of a tower. Its displacement in the first three seconds and in the last second is the same. Find the height of the tower.

Explanation

Let the hight of the tower be h and the time of journey be `t_(0)`.
Displacement in the first `3 s`,
`h_(1)=0+1/2g(3)^(2) …(i)`
displacement in the last second `t_(0)`
`h_(2)=u+1/2g(2t-1)`
`=0+1/2g(2 t_(0)-1) …(ii)`

Given `h_(1)=h_(2)`
`1/2 g(3)^(2)=1/2g (2 t_(0)-1)`
`g=2t_(0)-1impliest_(0)=5 s`
Time of journey `t_(0)=5 s`
For height of tower,
`h=ut+1/2g t^(2)`
`=0+1/2g t_(0)^(2)=1/2xx10xx(5)^(2)=125 m`