Two balls are projected simultaneously with the same speed from the top of a tower, one vertically upwards and the other vertically downwards. They reach the ground in 9 s and 4 s, respectively. The height of the tower is `(g=10 m//s^(2))`

Explanation

Let the balls be projected with speed u.
Treating `O` as the origin and vertically upward direction positive, displacement from `O` to `A=-h`
`t=t_(1)=9 s`
`h=ut-1/2 g t^(2)`
`-h=uxx9-1/2g(9)^(2) …(i)`
Treating `O` as the origin and vertically downward direction positive, displacement from `O` to `A=h`

`t=t_(2)=4 s`
`h=ut+1/2g t^(2)`
`h=uxx4+1/2g (4)^(2) ...(ii)`
From (i), `u=-h/4+1/2g(9)`
From (ii), `u=h/4-1/2 g (4)`
`-h/9+1/2g (9)=h/4-1/2g(4)`
`h/4+h/9=1/2g(4+9)`
`(13h)/36=5xx13impliesh=180m`
Height of the tower `=180m`