From the top of a tower, two balls are thrown horizontally with velocities `u_(1)` and `u_(2)` in opposite directions. If their velocities are perpendicular to each other just before they strike the ground, find the height of tower.

Explanation

Time taken to reach ground `t=sqrt(2h/g)`
at time of reaching ground respective velocities are `vecv_(1)=u_(1)hati+"gt"hatj, vecv_(2)=-u_(2)hati+"gt"hatj`
Given `vecv_(1).vecv_(2)=0, t=sqrt(u_(1)u_(2))/g`
`:. sqrt((2h)/(g))=sqrt(u_(1)u_(2))/g rArr h=(u_(1)u_(2))/(2g)`
is the height of the tower.