Two tall buildings face each other and are at a distance of 180 m from each other. With what velocity must a ball be thrown horizontally from a window 55 m above the ground in one building, so that it enters a window 10.9 m above the ground in the second building ?

Explanation

in figure P and Q are two tall buildings which are 180m apart.
`w_(1)` and `w_(2)` are the two windows in P and Q respectively.
Vertical downward distance to be covered by the ball
`="height of" W_(1)-"Height" W_(2)` ltbr. `=55-10=45m`
Initial vertical velocity of ball, `u_(y)=0`
`because y=u_(y)t+(1)/(2)"gt"^(2)`
`therefore 45=0+(1)/(2)xx10t^(2)` or `t=3s`.
Required horizontal velocity `=("Horizontal distance")/("Time")=(180)/(3)=60ms^(-1)`