Two particle move in a unifrom gravitational field with an acceleration "g". At the initial moment the paricles were located at one point and moved with velcoities `u_(1) = 3.0 ms^(-1)` and `u_(2) = 4.0 ms^(-1)` horizontally in opposite directions. Find the distance between the patricles at the moment when their their velocity vectors become mutually perpendicular .

Explanation

The situation is shown fig.

Let the velocity vectors perpendicular after time "t" when both particles has fallen same vertical distance `(1)/(2)gt^(2)` and have acquired same verticle velcoities gt.
Let the resultant velcoities makes angles `theta_(1)` and `theta_(2)` with horizontal.
Then `tan theta_(1) = ("gt")/(v_(1)) and tan theta_(2) = ("gt")/(v_(2))`
Sicne velcoity vectors are perpendicular
`theta_(1) + theta_(2) = 90^(@)` hence `theta_(2) = cot theta_(1)`
It makes `tan theta_(1) = ("gt")/(v_(1))` and `cotd theta_(1)=("gt")/(v_(2))`
on mulitplying we get ` t = (sqrt(v_(1)v_(2)))/(g) = and D =(u_(1) + u_(2))t`