Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located over a tower at one point and moved with velocities `v_1 = 3m//s and v_2= 4m//s` horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

Explanation

The situation is shown fig.

Let the velocity vectors become perpendicular after time t when both particles has fallen same vertical distance `1/2"gt"^(2)` and have acquired same vertical velocites gt.
Let their resultant velocities make angle `theta_(1)` and `theta_(2)` with horizontal.
Then `tan theta_(1)=("gt")/(v_(1))` and `tan theta_(2)=("gt")/(v_(2))`
Since velocity vectors are perpendicular
`theta_(1)+theta_(2)=90^(@)`, hence `tan theta_(2)=cot theta_(1)`
It makes `tan theta_(1)=("gt")/(v_(1))` and `cot theta_(1)=("gt")/(v_(2))`
On multiplying we get `t=(sqrt(v_(1)v_(2)))/g` and `D=(u_(1)+u_(2))t`