Two tall buildings face each other and are at a distance of `18 0 m` from wach other. With what velocity must a ball be thrown borixontall y from a window `55 m` above the fround in one building, so that it enters a window ` 10.9 m` above the ground in second window . `g=9.8 m//s^(-2)`

Explanation

Refer to Fig. 2 (d) . 19 , `A and B` respectively.
.
Vertical distanc ebetween two winows
`55 -10. 9 =44 .1 m`
Taking vertical downward motion of ball from `
`W_(1) to W_(2),`
we have , `u_(y) =0, a_(y) =g =9.8 ms^(-`, t=?`
` y=44.1 m`
As `y=u_(y) t+ 1/2 a_(y) t^(2)`
:. ` 44. 1 =0 + 1/2 xx 9.8 xx 9. 8 xx t&(2) or t_(2) = (44. 1 xx 2) /( 9.8) =9`
or ` t=3 s`
Required horizontal velcity of ball
`(horizontal distance )/(time taken) = (180)/3 = 60 ms ^(-1)`.